If $$x = {e^{y + {e^{y + .....{\text{to }}\infty }}{\text{ }}}}$$ then $$\frac{{dy}}{{dx}}$$ is :
A.
$$\frac{x}{{1 + x}}$$
B.
$$\frac{1}{x}$$
C.
$$\frac{{1 - x}}{x}$$
D.
none of these
Answer :
$$\frac{{1 - x}}{x}$$
Solution :
$$\eqalign{
& x = {e^{y + x}} \cr
& \Rightarrow \log \,x = y + x \cr
& \therefore \,\frac{1}{x} = \frac{{dy}}{{dx}} + 1 \cr} $$
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-