Differentiation Test 43

Let $$\phi (x)=\left ( x+tan^{-1} x\right )-\dfrac{x}{1+x^{2}}$$
$$\therefore $$  $$\displaystyle \phi'(\mathrm{x})=1+\frac{1}{1+\mathrm{x}^{2}}-\frac{1-\mathrm{x}^{2}}{(1+\mathrm{x}^{2})^{2}}=1+\frac{2\mathrm{x}^{2}}{(1+\mathrm{x}^{2})^{2}}>0 \ for$$ all $$\mathrm{x}$$
$$\therefore $$ f(x) is increasing in the domain 
$$\phi(0)=0$$
$$\therefore $$    $$\displaystyle \mathrm{x}+t\mathrm{a}\mathrm{n}^{-1}\mathrm{x}-\frac{\mathrm{x}}{1+\mathrm{x}^{2}}>0, \mathrm{x}>0$$ 
$$\Rightarrow \mathrm{f}(\mathrm{x})>g(\mathrm{x}), \mathrm{x}>0$$

Let $$f(x) =ax^2+bx+c$$
According to the  given condition $$a>0, b^2-4ac <0$$.........(i)
$$\therefore g(x) = ax^2+bx+c+2ax+b+2a=ax^2+(b+2a)x+(b+c+2a)$$
Now discriminant of $$g(x)$$ is $$D=(b+2a)^2-4a(2a+c+b)=b^2+4a^2+4ab-4ab-4ac-8a^2=(b^2-4ac)-4a^2< 0 $$ using (i)
Hence $$g(x)> 0$$ for any $$x \in R$$

Given $$h'\left( x \right) ={ \left[ f\left( x \right)  \right]  }^{ 2 }+{ \left[ g\left( x \right)  \right]  }^{ 2 }$$

We have,
$$f(x) = \sqrt{x - 1} + \sqrt{x + 24 - 10\sqrt{x - 1}}, 1 < x < 26$$

$$f(x)=-\displaystyle \frac {x^3}{3}+x^2 \sin 1.5 a-x \sin a\cdot \sin 2a-5 arc \sin (a^2-8a+17)$$
$$f(x)=\displaystyle \frac { -{ x }^{ 3 } }{ 3 } +{ x }^{ 2 }\sin { 6 } -x\sin { 4\sin { 8 } -5\sin ^{ -1 }{ \left( { \left( a-4 \right)  }^{ 2 }+1 \right)  }  } $$
$$f'(x)=-{ x }^{ 2 }+2x\sin { 6-\sin { 4.\sin { 8 }  }  }$$ .......( $$a=4 $$ considering the domain of inverse sine function) 
$$f'\left( \sin { \quad 8 }  \right) =-\sin ^{ 2 }{ 8 } +2\sin { 6 } \sin { 8 } -\sin { 4 } \sin { 8 } $$
$$=\sin { 8\left( -\sin { 8+2\sin { 6-\sin { 4 }  }  }  \right)  } $$
$$=-\sin { 8 } \left( \sin { 8 } +\sin { 4-2\sin { 6 }  }  \right) $$
$$=-\sin { 8\left( 2\sin { 6\cos { 2-2\sin { 6 }  }  }  \right)  } $$
$$=2\sin { 8 } \sin { 6\left( 1-\cos { 2 }  \right)  } $$
Now, 
$$sin 8 > 0 $$           (Since,$$  2\pi  <  8 < 3\pi $$ )
$$ sin 6 < 0 $$          (Since, $$\pi < 6 < 2\pi$$)
$$(1-\cos 2) > 0$$      
Hence, $$f'(\sin { 8 } )<0$$

$$h'(x) = 2f(x) \times f'(x) + 2 g(x) g'(x)$$

We have,
$$f(x) + f(y) + f(z) + f(x)f(y)f(z) = 14$$ for all $$x,\space y,\space z \in R \quad\quad ...(i)$$
Putting $$x = y = z = 0$$, we get,
$$3f(0) + \left\{f(0)\right\}^3 = 14$$
$$\left\{f(0)\right\}^3 + 3f(0) - 14 = 0$$
$$f(0) = 2.$$
Now, putting $$y = z = x$$ in $$(i)$$, we get
$$3f'(x) + 3\left\{f(x)\right\}^2f'(x) = 0 \quad$$ for all $$x \in R$$
$$\left\{\left\{f(x)\right\}^2 + 1\right\}f'(x) = 0 \quad$$ for all $$x \in R$$
$$ f'(x) = 0 \quad$$ for all $$x \in R$$

Let $$y=e^x+x$$
on differentiating w.r.t y,
$$1=(e^x+1)\dfrac {dx}{dy}\Rightarrow \dfrac {dx}{dy}=\dfrac {1}{1+e^x}$$
$$\therefore \left (\dfrac {dx}{dy}\right )_{x=log_e2}=\dfrac {1}{1+e^{log 2}}=\dfrac {1}{3}$$
or $$\left [\dfrac {d}{dx}(f^{-1}(x))\right ]_{x=log 2}=\dfrac {1}{3}$$

Given:

$$A: \displaystyle \frac {d}{dx} (x^2+y^2=4)$$ at $$(\sqrt 2, \sqrt 2))$$
$$2x+2y\displaystyle \frac { dy }{ dx } =0$$
$$\displaystyle \frac { dy }{ dx } =-\displaystyle \frac { x }{ y } =-\displaystyle \frac { \sqrt { 2 }  }{ \sqrt { 2 }  } =-1$$
$$ B: \displaystyle \frac {d}{dx}(\sin y+\sin x=\sin x\cdot \sin y)$$ at $$(\pi, \pi)$$
$$\cos { y\displaystyle \frac { dy }{ dx }  } +\cos { x } =\sin { x } \cos { y\displaystyle \frac { dy }{ dx } +\sin { y } \cos { x }  } $$
$$\displaystyle \frac { dy }{ dx } \left( \cos { y } -\sin { x } \cos { y }  \right) =\sin { y\cos { x-\cos { x }  }  } $$
$$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { \cos { x\left( \sin { y-1 }  \right)  }  }{ \cos { y\left( 1-\sin { x }  \right)  }  } $$
$$B=\displaystyle \frac { -1\left( 0-1 \right)  }{ -1\left( 1-0 \right)  } =-1$$
$$C: \displaystyle \frac {d}{dx}(2e^{xy}+e^xe^y-e^x=e^{xy+1})$$ at $$(1, 1)$$
$$\left[ 2{ e }^{ xy }\left( x\displaystyle \frac { dy }{ dx } +y \right)  \right] +{ e }^{ x }{ e }^{ y }\displaystyle \frac { dy }{ dx } +{ e }^{ y }{ e }^{ x }-{ e }^{ x }={ e }^{ xy+1 }\left( x\displaystyle \frac { dy }{ dx } +y \right) $$
$$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { y{ e }^{ xy+1 }-2y{ e }^{ xy }-{ e }^{ x+y }+{ e }^{ x } }{ 2x{ e }^{ xy }+{ e }^{ x+y }-x{ e }^{ xy+1 } } $$.......... since $${ e }^{ x }{ e }^{ y }={ e }^{ x+y }$$
$$C=\displaystyle \frac { { e }^{ 2 }-2{ e }^{ 1 }-{ e }^{ 2 }+{ e }^{ 1 } }{ 2{ e }^{ 1 }+{ e }^{ 2 }-{ e }^{ 2 } } $$
$$=\displaystyle \frac { -{ e }^{ 1 } }{ 2{ e }^{ 1 } } =-\displaystyle \frac { 1 }{ 2 } $$
Therefore $$A-B-C=\displaystyle \frac { 1 }{ 2 }$$