Question

If y= ${\cos ^{ - 1}}(\ln x)$ ,then value of $\dfrac{{dy}}{{dx}}$ is:
A.$ - \dfrac{1}{{x\sqrt {1 + \ln x} }}$
B.$\dfrac{1}{{x\sqrt {1 + {{(\ln x)}^2}} }}$
C.$ - \dfrac{1}{{x\sqrt {1 - {{(\ln x)}^2}} }}$
D. none of these.

Answer 1

Hint: In this question, we have ‘y’ on LHS and inverse function on RHS. So, we take cosine on both sides and then differentiate on both sides w.r.t. ‘x’. We will use chain rule of differentiation. Finally, rearrange and put the value of ‘y’ to get the answer.

Complete step-by-step answer:
The given expression is:
y= ${\cos ^{ - 1}}(\ln x)$
Since the term on RHS is an inverse function . So, we will take cosine on both sides.
On taking cosine on both sides, we get:
Cosy = lnx
On differentiating on both sides w.r.t. ‘x’, we get:
$ - \sin y\dfrac{{dy}}{{dx}} = \dfrac{1}{x}$
On solving the terms on RHS, we get:
$\dfrac{{dy}}{{dx}} = - \dfrac{1}{x} \times \dfrac{1}{{\sin y}}$ (1)
Now, we know that ${\cos ^2}y + {\sin ^2}y = 1$
Also , we can write:
$\sin y = \sqrt {1 - {{\cos }^2}y} $
Putting the value of cosy, we have:
$\sin y = \sqrt {1 - {{(\ln x)}^2}} $
Putting the value of ‘sin y’ in equation 1, we have:
$ \Rightarrow $ $\dfrac{{dy}}{{dx}} = - \dfrac{1}{x} \times \dfrac{1}{{\sqrt {1 - {{(\ln x)}^2}} }}$
Therefore the required answer is option C

Note:
In this type of question which involves inverse composite function terms on one side and ‘y’ on the other side, it is advisable to convert it into trigonometric function by taking cosine or sine or tangent function on both sides. You must remember some of the important rules of differentiation.
1.Product rule: used to find differentiation of two functions in multiplication.
$\dfrac{{d\left( {f(x)g(x)} \right)}}{{dx}} = \dfrac{{df(x)}}{{dx}}g(x) + \dfrac{{dg(x)}}{{dx}}f(x)$
2. chain rule: used to find differentiation of a composite function.
$\dfrac{{d(f(g(x))}}{{dx}} = f'(g(x))g'(x)$