Continuity and Differentiability Test - 10

Given that,

\(\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}\mathrm{x}^{2} \ln |\mathrm{x}| & \mathrm{x} \neq 0 \\ 0 & \mathrm{x}=0\end{array}\right.\)

\(f'(x)=\left\{\begin{array}{c}2 x \ln |x|+x^{2} \times \frac{1}{x} &x \neq 0 \\ 0 & x=0\end{array} ~~~=\left\{\begin{array}{c}\frac{2 \ln x}{\frac{1}{x}}+x & x \neq 0 \\ 0 & x=0\end{array}\right.\right.\)

\(\mathrm{f}^{\prime}(0)=\lim _{\mathrm{x} \rightarrow 0} \mathrm{f}^{\prime}(\mathrm{x})\)

\(=\lim _{x \rightarrow 0} \frac{2 \operatorname{lnx}}{\frac{1}{x}}+\mathrm{x}\) 

(It will become \(\frac{\infty}{\infty}\) form. So we will use here L'hospital rule)

We know that:

\(\lim _{x \rightarrow a} \frac{\mathrm{f}(\mathrm{x})}{g(x)}=\lim _{x \rightarrow a} \frac{\mathrm{f}^{\prime}(\mathrm{x})}{g^{\prime}(x)}\)

\(\mathrm{f}^{\prime}(0)=\lim _{\mathrm{x} \rightarrow 0} 2 \frac{\frac{1}{x}}{\frac{-1}{x^{2}}}+\mathrm{x}\)

\(\mathrm{f}^{\prime}(0)=\lim _{\mathrm{x} \rightarrow 0}-2 \mathrm{x}+\mathrm{x}\)

\(\mathrm{f'}(0)=0\)

Hence, the correct option is (A).

Given:

\(y+\sin ^{-1}\left(1-x^{2}\right)=e^{x}\)

\(y=e^{x}-\sin ^{-1}\left(1-x^{2}\right)\)

Differentiating above with respect to x, we get:

\(\frac{\mathrm{dy}}{\mathrm{dx}}=e^{x}-\frac{1}{\sqrt{1-\left(1-\mathrm{x}^{2}\right)^{2}}}(-2 \mathrm{x})\)

\(\frac{\mathrm{dy}}{\mathrm{dx}}=e^{\mathrm{x}}+\frac{2 \mathrm{x}}{\sqrt{1-\left(1-2 \mathrm{x}^{2}+\mathrm{x}^{4}\right)}}\)

\(\frac{\mathrm{dy}}{\mathrm{dx}}=e^{\mathrm{x}}+\frac{2 \mathrm{x}}{\sqrt{2 x^{2}-\mathrm{x}^{4}}}\)

\(\frac{\mathrm{dy}}{\mathrm{d} x}=\mathrm{e}^{\mathrm{x}}+\frac{2}{\sqrt{2-\mathrm{x}^{2}}}\)

Hence, the correct option is (B).

It is given that: 

\(y=e^{x+e^{x+e^{x+\cdots \infty}}}\)

\(\therefore y=e^{x+\left(e^{x+e^{x+} \cdots \infty}\right)}=e^{x+y}\)

Differentiating both sides with respect to x and using the chain rule, we get:

\(\frac{d y}{d x}=\frac{d}{d x} e^{x+y}\)

\(\Rightarrow \frac{d y}{d x}=e^{x+y} \frac{d}{d x}(x+y)\)

\(\Rightarrow \frac{d y}{d x}=y\left(1+\frac{d y}{d x}\right)\)

\(\Rightarrow \frac{d y}{d x}=y+y \frac{d y}{d x}\)

\(\Rightarrow(1-y) \frac{d y}{d x}=y\)

\(\Rightarrow \frac{d y}{d x}=\frac{y}{1-y}\)

Hence, the correct option is (C).

Given:

\(f(x)=x^{3}+3 x^{2}+3 x-7\)

We know that:

\(\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\)

Derivative of a constant, i.e.,

\(\frac{\mathrm{d}(\text { constant })}{\mathrm{dx}}=0\)

Therefore, 

\(\frac{d f(x)}{d x}=3 x^{2}+6 x+3\)

Putting \(x=2\) in above, we get:

\(\frac{\mathrm{d} \mathrm{f}(\mathrm{x})}{\mathrm{dx}}=3(2)^{2}+6(2)+3\)

\(=3(4)+12+3\)

\(=12+12+3\)

\(\frac{\mathrm{d} \mathrm{f}(\mathrm{x})}{\mathrm{dx}}=27\)

The value of \(\frac{\mathrm{d} \mathrm{f}(x)}{\mathrm{dx}}\) at \(\mathrm{x}=2\) is 27.

Hence, the correct option is (C).

Given:

\(y=3 t^{2}-4 t-3\)

Differentiating above with respect to t, we get:

\(\frac{\mathrm{dy}}{\mathrm{dt}}=6 \mathrm{t}-4\)

Similarly,

\(x=8 t+5\)

Differentiating above with respect to t, we get:

\(\frac{\mathrm{dx}}{\mathrm{dt}}=8\)

\(\frac{\mathrm{dx}}{\mathrm{dt}}=8\)

As we know:

\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}\)

Then, 

\(\frac{d y}{d x}=\frac{6 t-4}{8}\)

At t = 6, we get:

\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{(6 \times 6)-4}{8}\)

\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{32}{8}\)

\(\frac{\mathrm{dy}}{\mathrm{dx}}=4\)

Hence, the correct option is (A).

Given: 

\(y=\frac{12 \tan x-4 \tan ^{3} x}{9-27 \tan ^{2} x}\)  

\(\Rightarrow y=\frac{4\left(3 \tan x-\tan ^{3} x\right)}{9\left(1-3 \tan ^{2} x\right)}\)

We know that, 

\(\tan 3 x=\frac{3 \tan x-\tan ^{3} x}{1-3 \tan ^{2} x}\)

\(\Rightarrow y=\frac{4}{9} \tan 3 x\)

Differentiating both sides with respect to \(x\), we get:

\(\Rightarrow \frac{d y}{d x}=\frac{4}{9} \sec ^{2} 3 x \times 3\)

\(\Rightarrow \frac{d y}{d x}=\frac{4}{3} \sec ^{2} 3 x\)

Again, differentiating with respect to \(x\), we get:

\(\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{4}{3} \times 2 \times \sec 3 x \times \tan 3 x \times \sec 3 x \times 3\)

\(\Rightarrow \frac{d^{2} y}{d x^{2}}=8 \sec ^{2} 3 x \tan 3 x\)

Hence, the correct option is (A).

Given here: 

\(\lim _{x \rightarrow 0} \frac{\sqrt[p]{1+x}-1}{x}\)

Apply L-Hospital rule,

\(\lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}\)

Therefore,

\(\lim _{x \rightarrow 0} \frac{\sqrt[p]{1+x}-1}{x}\)

\(=\lim _{x \rightarrow 0} \frac{\frac{d}{d x}\left((1+x)^{\frac{1}{p}}-1\right)}{\left(\frac{d x}{d x}\right)}\)

\(=\lim _{x \rightarrow 0} \frac{1}{p} \times(1+x)^{\frac{1}{p}-1}\)

\(=\frac{1}{p}(1+0)^{\frac{1}{p}-1}\)

\(=\frac{1}{p}\)

Hence, the correct option is (B).

Given that: 

\(y^{2}+x^{2}+3 x+5=0\)

We know that:

\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}\)

Differentiating with respect to \(x\), we get:

\(2 y \frac{d y}{d x}+2 x+3(1)+0=0\)

\(2 y \frac{d y}{d x}+2 x+3=0\)

\(2 y \frac{d y}{d x}=-(2 x+3)\)

\(\frac{d y}{d x}=-\frac{2 x+3}{2 y}\)

Now at (0, -3)

\(\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{2(0)+3}{2(-3)}\)

\(\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{3}{(-6)}\)

\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}\)

\(\frac{\mathrm{dy}}{\mathrm{dx}}=0.5\)

Hence, the correct option is (D).

We know that:

If \(f(x)=x^{n}\), then \(f'(x)=n x^{n-1}\) 

\(f(x)=\log x\), then \(f'(x)=\frac{1}{x}\) 

\(f(x)=\) Constant, then \(f'(x)=0\) 

\(\log m^{n}=n \log m\)

Given here:

\(\mathrm{f}(\mathrm{x})=\log \mathrm{x}^{3}+2 \mathrm{x}^{2}-3 \mathrm{x}\)

\(\Rightarrow \mathrm{f}(\mathrm{x})=3 \log \mathrm{x}+2 \mathrm{x}^{2}-3 \mathrm{x}\)

\(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=3 \frac{\mathrm{d}}{\mathrm{dx}} \log \mathrm{x}+\frac{\mathrm{d}}{\mathrm{dx}} 2 \mathrm{x}^{2}-\frac{\mathrm{d}}{\mathrm{dx}} 3 \mathrm{x}\)

\(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{3}{x}+4 \mathrm{x}-3\)

Putting x = 3 in above, we get:

\(\Rightarrow \mathrm{f}^{\prime}(3)=\frac{3}{3}+4(3)-3\)

\(\Rightarrow \mathrm{f}^{\prime}(3)=10\)

Hence, the correct option is (D).

Given here:

\(\lim _{x \rightarrow 0} \frac{\sqrt{4+x}-2}{3 x}\)

On substituting the limit value in the given function, we get:

\(\lim _{x \rightarrow 0} \frac{\sqrt{4+x}-2}{3 x}\)

\(=\frac{\sqrt{(4+0)}-2}{3(0)}\)

\(=\frac{0}{0}\)

Conjugate of \(\sqrt{4+x}-2\) is \(\sqrt{4+x}+2\)

Multiply and divide by \(\sqrt{4+x}+2\) in the given expression, we get:

\(\therefore \lim _{x \rightarrow 0} \frac{\sqrt{4+x}-2}{3 x} \times \frac{\sqrt{4+x}+2}{\sqrt{4+x}+2}\)

\(=\lim _{x \rightarrow 0} \frac{4+x-4}{3 x \sqrt{4+x}+2} \quad\left(\because(a+b)(a-b)=\left(a^{2}-b^{2}\right)\right)\)

\(= \lim _{x \rightarrow 0} \frac{x}{3 x(\sqrt{4+x}+2)}\)

\(=\lim _{x \rightarrow 0} \frac{1}{3(\sqrt{4+x}+2)}\)

Now, putting the value of x in the limit, we get:

\(=\frac{1}{3(2+2)}\)

\(=\frac{1}{12}\)

Hence, the correct option is (B).

Since \(\mathrm{f}(\mathrm{x})\) is given to be continuous at \(\mathrm{x}=0\), 

\(\lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})=\mathrm{f}(0)\)

Also, \(\lim _{x \rightarrow a^{+}} f(x)=\lim _{x \rightarrow a^{-}} f(x)\) because \(f(x)\) is same for \(x>0\) and \(x<0\).

\(\therefore \lim _{x \rightarrow 0} f(x)=f(0)\)

We know that:

\(\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\)

\(\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=1\)

Therefore,

\(\Rightarrow \lim _{x \rightarrow 0} \frac{\sin 3 x}{e^{2 x}-1}=k-2\)

Multiplying and dividing by 3x in numerator and by 2x in denominator, we get:

\(\Rightarrow \lim _{x \rightarrow 0} \frac{\frac{\sin 3 x}{3 x} \times 3 x}{\frac{e^{2 x}-1}{2 x} \times 2 x}=k-2\)

\(\Rightarrow \frac{3}{2}=k-2\)

\(\Rightarrow k=\frac{7}{2}\)

Hence, the correct option is (D).

Given that: 

\(\lim _{\mathrm{x} \rightarrow 0} \frac{\log (1+\sin x)}{\mathrm{x}}=\mathrm{k}\)

Put x = 0, to check form

\(\lim _{x \rightarrow 0} \frac{\log (1+\sin x)}{x}\)

\(=\frac{\log (1+0)}{0}\)

\(=\frac{0}{0}\)

Applying L’ hospital’s rule as,

\(\lim _{x \rightarrow a} \frac{\mathrm{f}(x)}{g(x)}=\lim _{x \rightarrow a} \frac{\mathrm{f}^{\prime}(x)}{g^{\prime}(x)}\)

\(\lim _{x \rightarrow 0} \frac{\log (1+\sin x)}{x}=\lim _{x \rightarrow 0} \frac{\frac{d}{d x}(\log (1+\sin x))}{\frac{d}{d x}(x)}\)

\(=\lim _{\mathbf{x} \rightarrow 0} \frac{\frac{1}{1+\sin x} \times \cos x}{1}\)

\(=\frac{1}{1+\sin 0} \times \cos 0\)

\(=1\)

\(\therefore \mathrm{k}=1\)

Hence, the correct option is (D).

We know that:

If \(f(x)=x^{n}\), then: 

\(f^{\prime}(x)=n x^{n-1}\)

Given here:

\(\lim _{x \rightarrow 1} \frac{4 x^{2}-5 x+1}{x^{2}-1}\)

 We can see here:

\(\lim _{x \rightarrow 1} \frac{4 x^{2}-5 x+1}{x^{2}-1}=\frac{0}{0}\), is an indeterminate form.

Let us use the L'Hospital's rule:

\(\lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}\)

Then:

\(\lim _{x \rightarrow 1} \frac{4 x^{2}-5 x+1}{x^{2}-1}\)

\(=\lim _{x \rightarrow 1} \frac{\frac{d}{d x}\left(4 x^{2}-5 x+1\right)}{\frac{d}{d x}\left(x^{2}-1\right)}\)

\(=\lim _{x \rightarrow 1} \frac{8 x-5}{2 x}\)

\(=\frac{3}{2}\)

Hence, the correct option is (C).

Given: 

\(\sqrt{y}=\sin x+\cos x\)  

Squaring both sides, we get:

\(\Rightarrow y=(\sin x+\cos x)^{2}\)

\(\Rightarrow y=\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x\)

\(\Rightarrow y=1+\sin 2 x \quad(\because 2 \sin x \cos x=\sin 2 x)\)

Differentiating above with respect to x, we get:

\(\frac{\mathrm{dy}}{\mathrm{dx}}=0+2 \times \cos 2 \mathrm{x}\)

\(=2 \cos 2 \mathrm{x}\)

Hence, the correct option is (B).

Given:

\(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{3}+\frac{1}{\mathrm{x}}\)

We know that:

If \(f(x)=x^{n}\), then: 

\(f'(x)=n x^{n-1}\)

Differentiating with respect to x, we get:

\(f'(x)=6 x^{2}-\frac{1}{x^{2}}\)

Putting \(x=1\) in above, we get:

\(f'(1)=6 \times 1^{2}-\frac{1}{1^{2}}\)

\(f'(1)=6-1=5\)

\(\therefore\) The value of \(f^{\prime}(1)\) is 5.

Hence, the correct option is (A).

Given:

\(\lim _{x \rightarrow 0} \frac{\sin x \log (1-x)}{x^{2}}\)

As we know,

\(\lim _{x \rightarrow a}[f(x) \cdot g(x)]=\lim _{x \rightarrow a} f(x) \cdot \lim _{x \rightarrow a} g(x)\)

\(\therefore \lim _{x \rightarrow 0} \frac{\sin x \log (1-x)}{x^{2}}\)

\(=\lim _{x \rightarrow 0} \frac{\sin x}{x} \times \lim _{x \rightarrow 0} \frac{\log (1-x)}{x}\)

\(=1 \times \lim _{x \rightarrow 0} \frac{\log (1+(-x))}{x} \quad\) \( (\because\lim _{x \rightarrow 0} \frac{\sin x}{x}=1)\)

\(=\lim _{x \rightarrow 0} \frac{\log (1+(-x))}{-(-x)}\)

\(=-1 \times \lim _{x \rightarrow 0} \frac{\log (1+(-x))}{(-x)}\) 

\(=-1 \times 1 \quad\) \((\because\lim _{x \rightarrow 0} \frac{\log x}{x}=1)\)  

\(=-1\)

Hence, the correct option is (A).

Let, \(y=\frac{1}{3 x^{2}}\) ....(1)

We know that:

\(\frac{d x^{n}}{dx}=nx^{n-1}\)

Differentiating (1) with respect of x, we get:

\(\frac{dy}{dx} =\frac{d}{dx}\left(\frac{1}{3 x^{2}}\right)\)

\(=\frac{1}{3} \frac{d\left(x^{-2}\right)}{dx}\)

\(=\frac{1}{3} \times-2 \times x^{-2-1}\)

\(=\frac{-2}{3 x^{3}}\)

Hence, the correct option is (C).

Given:

\(\lim _{x \rightarrow 0} \frac{3^{x}+3^{-x}-2}{x}\)

We know that:

\(\lim _{x \rightarrow a}[f(x)+g(x)]=\lim _{x \rightarrow a} f(x)+\lim _{x \rightarrow a} g(x)\)

\(\lim _{x \rightarrow 0} \frac{\left(a^{x}-1\right)}{x}=\log a\)

\(\log m^{n}=n \log m\)

Therefore,

\(\lim _{x \rightarrow 0} \frac{3^{x}+3^{-x}-2}{x}\)

\(=\lim _{x \rightarrow 0} \frac{3^{x}-1+3^{-x}-1}{x}\)

\(=\lim _{x \rightarrow 0} \frac{3^{x}-1}{x}+\lim _{x \rightarrow 0} \frac{3^{-x}-1}{x}\)

\(=\lim _{x \rightarrow 0} \frac{3^{x}-1}{x}+\lim _{x \rightarrow 0} \frac{\left(3^{-1}\right)^{x}-1}{x}\)

\(=\log 3+\log \left(3^{-1}\right)\)

\(=\log 3-\log 3\)

\(=0\)

Hence, the correct option is (A).

Given:

\(\lim _{x \rightarrow 2} \frac{\sin \left(e^{x-2}-1\right)}{\log (x-1)}\)  .....(1)

We know that:

\(\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=l=\lim _{x \rightarrow a} f(x)\)

On substituting, \(\mathrm{h}=\mathrm{x}-2\) in (1), we get:

\(=\lim _{h \rightarrow 0} \frac{\sin \left(e^{h}-1\right)}{\log (1+h)}\)

This can be rearranged as:

\(=\lim _{h \rightarrow 0} \frac{\sin \left(e^{h}-1\right)}{e^{h}-1} \cdot \frac{e^{h}-1}{h} \cdot \frac{h}{\log (1+h)}\)

\(=1 \cdot 1 \cdot 1\)

\(=1\) 

And, \(f(2)=k\)

∴ For the function to be continuous the value of the function f(x) at x = 2 must equal the limiting value of 1, i.e., k = 1

Hence, the correct option is (D).

Let, \(v=\csc ^{-1}\left(\frac{1}{2 x^{2}-1}\right)\) and \(u=\sqrt{1-x^{2}}\)

Substituting \(x=\cos \theta\), 

\(\Rightarrow \theta=\cos ^{-1} x\), 

We get:

\(v=\csc ^{-1}\left(\frac{1}{2 \cos ^{2} \theta-1}\right)\)

\(=\csc ^{-1}\left(\frac{1}{\cos ^{2} \theta-\sin ^{2} \theta}\right)\)

\(=\csc ^{-1}\left(\frac{1}{\cos 2 \theta}\right)\)

\(=\csc ^{-1}(\sec 2 \theta)\)

\(=\csc ^{-1}\left(\csc \left(90^{\circ}-2 \theta\right)\right)\)

\(=90^{\circ}-\) \(2 \theta\).

And, \(u=\sqrt{1-x^{2}}\)

\(=\sqrt{1-\cos ^{2} \theta}\)

\(=\sqrt{\sin ^{2} \theta}\)

\(=\sin \theta\)

Now, 

\(\frac{d v}{d \theta}=-2\) 

And, 

\(\frac{d u}{d \theta}=\cos \theta\)

\(\therefore \frac{\mathrm{dv}}{\mathrm{du}} =\frac{\mathrm{dv}}{\mathrm{d} \theta} \times \frac{\mathrm{d} \theta}{\mathrm{d} u}\)

\(=\frac{-2}{\cos \theta}\)

\(=\frac{-2}{x}\)

Hence, the correct option is (B).

Given that:

\(\mathrm{y}=\tan ^{-1}\left[\frac{8 \mathrm{x}}{1-15 \mathrm{x}^{2}}\right]\)

As we know that, 

\(\tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{y}=\tan ^{-1}\left[\frac{x+\mathrm{y}}{1-\mathrm{xy}}\right]\)

So, 

\(y=\tan ^{-1}\left[\frac{5 x+3 x}{1-5 x \cdot 3 x}\right]\)

\(=\tan ^{-1} 5 x+\tan ^{-1} x\)

Also, we know:

\(\frac{\mathrm{d}\left(\tan ^{-1} \mathrm{x}\right)}{\mathrm{dx}}=\frac{1}{1+\mathrm{x}^{2}}\)

Differentiating y with respect to x, we get:

\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}\left(\tan ^{-1} 5 \mathrm{x}\right)}{\mathrm{dx}}+\frac{\mathrm{d}\left(\tan ^{-1} 3 \mathrm{x}\right)}{\mathrm{dx}}\)

\(=\frac{5}{1+(5 x)^{2}}+\frac{3}{1+(3 \mathrm{x})^{2}}\)

\(=\frac{5}{1+25 x^{2}}+\frac{3}{1+9 x^{2}}\)

Hence, the correct option is (B).

Given:

\(\frac{\mathrm{d}}{\mathrm{dx}}\left(\cot ^{-1} \frac{1}{x}\right)\)

\(=\frac{\mathrm{d} \cot ^{-1} \frac{1}{\mathrm{x}}}{\mathrm{d}\left(\frac{1}{x}\right)} \times \frac{\mathrm{d}\left(\frac{1}{x}\right)}{\mathrm{dx}}\)

We know that:

\(\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}}\)

\(\frac{d}{d x}\left(\cot ^{-1} x\right)=\frac{-1}{1+x^{2}}\)

\(\frac{d}{d x}\left(\frac{1}{x}\right)=-\frac{1}{x^{2}}\)

\(=\frac{-1}{1+\left(\frac{1}{x}\right)^{2}} \times \frac{d}{d x}\left(\frac{1}{x}\right)\)

Therefore, 

\(\frac{\mathrm{d} \cot ^{-1} \frac{1}{\mathrm{x}}}{\mathrm{d}\left(\frac{1}{x}\right)} \times \frac{\mathrm{d}\left(\frac{1}{x}\right)}{\mathrm{dx}}\)

\(=\frac{-1}{1+\left(\frac{1}{x}\right)^{2}} \times\left(-\frac{1}{x^{2}}\right)\)

\(=\frac{1}{\frac{x^{2}+1}{x^{2}}} \times\left(\frac{1}{x^{2}}\right)\)

\(=\frac{x^{2}}{1+x^{2}} \times \frac{1}{x^{2}}\)

\(=\frac{1}{1+x^{2}}\)

Hence, the correct option is (C).

Given:

\(f(x)=4 x^{3}-3 x^{2}+5\)

We know that:

If \(f(x)=x^{n}\), then: 

\(f'(x)=n x^{-1}\)

Differentiating above with respect to x, we get:

\(f'(x)=4 \times 3 \times x^{2}-3 \times 2 \times x + 0\)

\(f'(x)=12 x^{2}-6 x\)

Again, differentiating above with respect to x, we get:

\(\Rightarrow f''(x)=24 x-6\)

Hence, the correct option is (B).

Given:

\(\frac{\mathrm{d}^{2} \cot ^{-1} \mathrm{x}}{\mathrm{d} x^{2}}\)

\(=\frac{\mathrm{d}}{\mathrm{dx}} \times \frac{\mathrm{d} (\cot ^{-1} \mathrm{x})}{\mathrm{dx}}\)

\(=\frac{\mathrm{d}\left(\frac{-1}{1+\mathrm{x}^{2}}\right)}{\mathrm{d} x}\)

\(=-\frac{\mathrm{d}\left(\frac{1}{1+x^{2}}\right)}{\mathrm{dx}}\)

On applying chain rule of derivative, we get:

\(=-\frac{\mathrm{d}\left(\frac{1}{1+x^{2}}\right)}{\mathrm{d}\left(1+\mathrm{x}^{2}\right)} \times \frac{\mathrm{d}\left(1+\mathrm{x}^{2}\right)}{\mathrm{dx}}\)

\(=\frac{1}{\left(1+\mathrm{x}^{2}\right)^{2}} \times(0+2 \mathrm{x}) \quad\left(\because \frac{\mathrm{d}\left(\frac{1}{\mathrm{x}}\right)}{\mathrm{dx}}=\frac{-1}{\mathrm{x}^{2}}\right)\)

\(=\frac{2 \mathrm{x}}{\left(1+\mathrm{x}^{2}\right)^{2}}\)

Hence, the correct option is (D).

Given here, 

\(f(x)=x-\frac{1}{x}\) ....(1)

We know that:

If \(f(x)=x^{n}\), then: 

\(f^{\prime}(x)=n x^{n-1}\)

Differentiating (1) with respect to x, we get:

\(\Rightarrow f^{\prime}(x)=1-\left(\frac{-1}{x^{2}}\right)\)

\(f^{\prime}(x)=1+\frac{1}{x^{2}}\)

Putting x = -1 in above, we get:

\(f'(-1)=1+\frac{1}{(-1)^{2}}\)

\(=1+1\)

\(=2\)

Hence, the correct option is (B).

Given here: 

\(f(2)=2\) and \(f(x)=2\).

Let us check the form by putting \(x = 2\), we get:

\(\lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}\)

\(=\frac{2(2)-2(2)}{2-2}\)

\(=\frac{0}{0}\)

Apply L’ hospital’s rule,

\(\lim _{x \rightarrow a} \frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(x)}=\lim _{\mathrm{x} \rightarrow \mathrm{a}} \frac{\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{f}(\mathrm{x}))}{\frac{d}{d x}(\mathrm{~g}(\mathrm{x}))}=\lim _{\mathrm{x} \rightarrow \mathrm{a}} \frac{\mathrm{f}^{\prime}(x)}{\mathrm{g}^{\prime}(x)}\)

\(\lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}\)

Differentiating the numerator and the denominator in above, we get:

\(=\lim _{x \rightarrow 2} \frac{\frac{d}{d x}(x f(2)-2 f(x))}{\frac{d}{d x}(x-2)}\)

\(=\lim _{x \rightarrow 2} \frac{f(2)-2 f^{\prime}(x)}{1}\)

Now, take limit at x = 2

\(\lim _{x \rightarrow 2}=f(2)-2 f^{\prime}(2)\)

\(=2-2(2)\)

\(=-2\)

Hence, the correct option is (C).

Given:

\(y=x^{x}\)

Taking log both sides, we get,

\(\Rightarrow \log y=\log x^{x} \)

\(\Rightarrow \log y=x \log x\quad\left(\because \log m^{n}=n \log m\right)\)

Differentiating above with respect to x, we get:

\(\Rightarrow \frac{1}{y} \times \frac{d y}{d x}=x \times \frac{\log x}{d x}+\log x \times \frac{d x}{d x}\)

\(\Rightarrow \frac{1}{y} \times \frac{d y}{d x}=x \times \frac{1}{x}+\log x \times 1\)

\(\Rightarrow \frac{d y}{d x}=y \times[1+\log x]\)

\(\Rightarrow \frac{d y}{d x}=x^{x} \times[1+\log x]\)

Putting x = 1, we get:

\(\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=1^{1} \times[1+\log 1]\)

\(\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=1 \times[1+0] \quad(\because \log 1=0)\)

\(\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=1\)

Hence, the correct option is (B).

Given:

\(f(x)=2 x^{2}+\frac{1}{x}\) .... (1)

We know that:

If \(f(x)=x^{n}\), then: 

\(f'(x)=n x^{n-1}\)

Differentiating (1) with respect to x, we get:

\(\Rightarrow f^{\prime}(x)=4 x-\frac{1}{x^{2}}\)

Putting x = 1 in above, we get:

\(f'(1)=4 \times 1-\frac{1}{1^{2}}\)

\(f'(1)=4-1=3\)

\(\therefore\) The value of \(f^{\prime}(1)\) is 3.

Hence, the correct option is (C).

Given that: 

\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)

Differentiating with respect to \(x\), we get: 

\(\Rightarrow \frac{2 {x}}{{a}^{2}}+\frac{2 {y}}{{b}^{2}} \frac{{dy}}{{dx}}=0\)

\(\Rightarrow \frac{2 {y}}{{b}^{2}} \frac{{dy}}{{dx}}=-\frac{2 {x}}{a^{2}}\)

\(\therefore \frac{{dy}}{{dx}}=-\frac{{b}^{2} {x}}{{a}^{2} {y}}\)

Hence, the correct option is (B).

Given:

\(x^{e}=e^{x^{2}+y^{2}}\)

Taking log on both sides, we get:

\(\Rightarrow \log x^{e}=\log e^{x^{2}+y^{2}}\)

\(\Rightarrow e \log x=\left(x^{2}+y^{2}\right) \log e\)

\(\left[\because \log m^{n}=n \log m\right]\)

\(\Rightarrow e \log x=x^{2}+y^{2} \quad [\because \log e=1]\)

Differentiating with respect to x, we get:

\(\Rightarrow \mathrm{e}\left(\frac{1}{\mathrm{x}}\right)=2 \mathrm{x}+2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}\)

\(\Rightarrow \frac{\mathrm{e}}{\mathrm{x}}-2 \mathrm{x}=2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}\)

\(\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{e}-2 \mathrm{x}^{2}}{2 \mathrm{xy}}\)

Hence, the correct option is (B).