Calculus Test - 3

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,

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

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

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

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

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

We have,

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

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

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

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

\(=f(2)-2 f^{\prime}(2) \) \(\quad{\left[\because f^{\prime}(2)=\lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x-2}\right]}\)

\(=4-2 \times 1=2\) \( \quad {\left[\because f(2)=4 \text { and } f^{\prime}(2)=1\right]}\)

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:

\(f'(x)=4 \times3 \times x^{2}-3 \times2 \times x + 0\)

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

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

(x)=24 x-6)

Given:

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

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

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

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

On applying chain rule of derivative, we get:

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

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

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

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:

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

\(=1+1\)

\(=2\)

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:

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

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

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

The given function is a constant function for all \(x<3\) and for all \(x>5\) so it is continuous for all \(x<3\) and for all \(x>5\). We know that a polynomial function is continuous. So, the given function is continuous for all \(x \in(3,5)\). 

Thus, \(f(x)\) is continuous at each \(x \in R\) except possibly at \(x=3\) and \(x=5\).

At, \(x=3\), we have

\(\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3} 1=1, \lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3} a x+b=3 a+b\) and, \(f(3)=1\)

For \(f(x)\) be continuous at \(x=3\), we must have

\( \lim _{x \rightarrow 3^{-}} f(x)  =\lim _{x \rightarrow 3^{+}} f(x)=f(3) \)

\(\Rightarrow  1 =3 a+b\cdots(i)\)

At \(x=5\), we have

\(\lim _{x \rightarrow 5^{-}} f(x)=\lim _{x \rightarrow 5} a x+b=5 a+b ; \lim _{x \rightarrow 5^{-}} f(x)=\lim _{x \rightarrow 5} 7=7\) and  \(f(5)=7\)

For \(f(x)\) to be continuous at \(x=5\), we must have

\(\lim _{x \rightarrow 5^{-}} f(x)=\lim _{x \rightarrow 5^{+}} f(x)=f(5) \)

\(\Rightarrow  5 a+b  =7\cdots(ii)\)

Solving (i) and (ii), we get \(a=3, b=-8\)

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

Given that:

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

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

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

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

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

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:

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

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

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