If $$t\left( {1 + {x^2}} \right) = x$$ and $${x^2} + {t^2} = y$$ then $$\frac{{dy}}{{dx}}$$ at $$x=2$$ is :
A.
$$\frac{{88}}{{125}}$$
B.
$$\frac{{488}}{{125}}$$
C.
1
D.
none of these
Answer :
$$\frac{{488}}{{125}}$$
Solution :
$$\eqalign{
& {x^2} + {t^2} = y \cr
& \Rightarrow \frac{{dy}}{{dx}} = 2x + 2t.\frac{{dt}}{{dx}}......\left( 1 \right) \cr
& {\text{As }}t = \frac{x}{{1 + {x^2}}} \Rightarrow \frac{{dt}}{{dx}} = \frac{{1 - {x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}} \cr
& {\text{Substitute these value of }}t{\text{ and }}\frac{{dt}}{{dx}}{\text{ in }}\left( 1 \right){\text{, we get}} \cr
& \frac{{dy}}{{dx}} = 2x + \frac{{2x}}{{1 + {x^2}}}.\frac{{1 - {x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}} \cr
& {\text{On putting }}x = 2{\text{ in }}\frac{{dy}}{{dx}},{\text{ we get}} \cr
& \frac{{dy}}{{dx}} = \frac{{488}}{{125}} \cr} $$
Releted MCQ Question on Calculus >> Differentiability and Differentiation
Releted Question 1
There exist a function $$f\left( x \right),$$ satisfying $$f\left( 0 \right) = 1,\,f'\left( 0 \right) = - 1,\,f\left( x \right) > 0$$ for all $$x,$$ and-
A.
$$f''\left( x \right) > 0$$ for all $$x$$
B.
$$ - 1 < f''\left( x \right) < 0$$ for all $$x$$
C.
$$ - 2 \leqslant f''\left( x \right) \leqslant - 1$$ for all $$x$$
If $$f\left( a \right) = 2,\,f'\left( a \right) = 1,\,g\left( a \right) = - 1,\,g'\left( a \right) = 2,$$ then the value of $$\mathop {\lim }\limits_{x \to a} \frac{{g\left( x \right)f\left( a \right) - g\left( a \right)f\left( x \right)}}{{x - a}}$$ is-
Let $$f:R \to R$$ be a differentiable function and $$f\left( 1 \right) = 4.$$ Then the value of $$\mathop {\lim }\limits_{x \to 1} \int\limits_4^{f\left( x \right)} {\frac{{2t}}{{x - 1}}} dt$$ is-