Question

If $f\left( x \right) = {e^x}$, $g\left( x \right) = {\sin ^{ - 1}}x$ and $h\left( x \right) = f\left[ {g\left( x \right)} \right]$, then $\dfrac{{h'\left( x \right)}}{{h\left( x \right)}}$ is equal to:
(A) ${e^{{{\sin }^{ - 1}}x}}$
(B) \[\dfrac{1}{{\sqrt {1 - {x^2}} }}\]
(C) ${\sin ^{ - 1}}x$
(D) $\dfrac{1}{{1 - {x^2}}}$

Answer 1

Hint: In the given question, we are required to find the value of an expression. We are given two functions in the problem and first we have to calculate the composite function of the two. Then, we use the rules of differentiation such as the chain rule to evaluate the expression $\dfrac{{h'\left( x \right)}}{{h\left( x \right)}}$. A thorough understanding of functions and applications of chain rule can be of great significance.

Complete step-by-step answer:
In the given question, we are given a functions $f\left( x \right) = {e^x}$, $g\left( x \right) = {\sin ^{ - 1}}x$ and $h\left( x \right) = f\left[ {g\left( x \right)} \right]$.
Now, we have to find a composite function using the given original functions.
So, $h\left( x \right) = f\left[ {{{\sin }^{ - 1}}x} \right]$
So, we have to change the value of the variable in the original function in order to find the composite function.
Originally, $f\left( x \right) = {e^x}$. So, we get,
\[h\left( x \right) = f\left[ {{{\sin }^{ - 1}}x} \right] = {e^{{{\sin }^{ - 1}}x}}\]
Now, we have to evaluate the expression $\dfrac{{h'\left( x \right)}}{{h\left( x \right)}}$. So, we need to differentiate the function $h\left( x \right) = {e^{{{\sin }^{ - 1}}x}}$ with respect to x using the chain rule of differentiation.
So, $\dfrac{d}{{dx}}\left[ {h\left( x \right)} \right] = \dfrac{d}{{dx}}\left[ {{e^{{{\sin }^{ - 1}}x}}} \right]$
Let $t = {e^{{{\sin }^{ - 1}}x}}$. So, we get,
\[ \Rightarrow h'\left( x \right) = \dfrac{d}{{dx}}\left[ {{e^t}} \right]\]
Now, we know that the derivative of ${e^x}$ with respect to x is ${e^x}$.
\[ \Rightarrow h'\left( x \right) = {e^t}\dfrac{{dt}}{{dx}}\]
Putting in the value of t in the expression, we get,
\[ \Rightarrow h'\left( x \right) = {e^{{{\sin }^{ - 1}}x}} \times \dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right)\]
Now, we know that the derivative of sine inverse functions as \[\dfrac{1}{{\sqrt {1 - {x^2}} }}\]. So, we get,
\[ \Rightarrow h'\left( x \right) = \dfrac{{{e^{{{\sin }^{ - 1}}x}}}}{{\sqrt {1 - {x^2}} }}\]
Now, $\dfrac{{h'\left( x \right)}}{{h\left( x \right)}} = \dfrac{{\left( {\dfrac{{{e^{{{\sin }^{ - 1}}x}}}}{{\sqrt {1 - {x^2}} }}} \right)}}{{{e^{{{\sin }^{ - 1}}x}}}} = \dfrac{1}{{\sqrt {1 - {x^2}} }}$.
So, we get the value of $\dfrac{{h'\left( x \right)}}{{h\left( x \right)}}$ as \[\dfrac{1}{{\sqrt {1 - {x^2}} }}\].
Hence, option (B) is the correct answer.
So, the correct answer is “Option B”.

Note: Such questions that require just simple change of variable can be solved easily by keeping in mind the algebraic rules such as substitution and transposition. Substitution of a variable involves putting a certain value in place of the variable. That specified value may be a certain number or even any other variable. We must know the chain rule of differentiation to solve such problems.