Question

Let $f\left( x \right) = {e^x}$, $g\left( x \right) = {\sin ^{ - 1}}x$ and $h\left( x \right) = f\left( {g\left( x \right)} \right)$ then find $\dfrac{{h'\left( x \right)}}{{h\left( x \right)}}$.

Answer 1

Here we will substitute the given function in the given relation and simplify it to find the value of $h\left( x \right)$. We will then differentiate it with respect to the variable $x$ using the differentiation formula to find the derivative of $h\left( x \right)$. We will then divide the derivative of $h\left( x \right)$ by $h\left( x \right)$ and simplify further to get the required value.

Complete step by step solution:
The values of two functions are given as
$f\left( x \right) = {e^x}$……………………..$\left( 1 \right)$
$g\left( x \right) = {\sin ^{ - 1}}x$………………..$\left( 2 \right)$
Here $f\left( x \right)$ is the composite function and $g\left( x \right)$ is the first function.
Now, as it is given that
$h\left( x \right) = f\left( {g\left( x \right)} \right)$
Rewriting the above equation, we get
$\Rightarrow h\left( x \right) = \left( {f \circ g} \right)\left( x \right)$
Substituting the value from equation $\left( 2 \right)$ in above equation we get,
$\Rightarrow h\left( x \right) = f\left( {{{\sin }^{ - 1}}x} \right)$
Now we will substitute ${\sin ^{ - 1}}x$ in place of $x$ in equation $\left( 1 \right)$. Therefore, we get
$f\left( {{{\sin }^{ - 1}}x} \right) = {e^{{{\sin }^{ - 1}}x}}$
Now substituting $f\left( {{{\sin }^{ - 1}}x} \right) = {e^{{{\sin }^{ - 1}}x}}$ in the equation $h\left( x \right) = f\left( {{{\sin }^{ - 1}}x} \right)$, we get
$\Rightarrow h\left( x \right) = {e^{{{\sin }^{ - 1}}x}}$………………………………..$\left( 3 \right)$
Now, the second value that we have to find is the derivative of $h\left( x \right)$.
So, differentiating $h\left( x \right)$ with respect to $x$, we get
$\Rightarrow h'\left( x \right) = \dfrac{{d\left( {{e^{{{\sin }^{ - 1}}x}}} \right)}}{{dx}}$…………………………………..$\left( 4 \right)$
Now, as we know differentiation of ${e^x}$ is calculated as
\dfrac{{d\left( {{e^x}} \right)}}{{dx}} = {e^x} \times \dfrac{{d\left( x \right)}}{{dx}} \\\ \Rightarrow \dfrac{{d\left( {{e^x}} \right)}}{{dx}} = {e^x} \\\
Similarly, we will find the derivative in equation $\left( 4 \right)$ by using above technique
$h'\left( x \right) = {e^{{{\sin }^{ - 1}}x}} \times \dfrac{{d\left( {{{\sin }^{ - 1}}x} \right)}}{{dx}}$
Differentiation of ${\sin ^{ - 1}}x = \dfrac{1}{{\sqrt {1 - {x^2}} }}$
So, we get
$\Rightarrow h'\left( x \right) = {e^{{{\sin }^{ - 1}}x}} \times \dfrac{1}{{\sqrt {1 - {x^2}} }}$……………………$\left( 5 \right)$

Now, dividing equation $\left( 5 \right)$ by equation $\left( 3 \right)$, we get
$\dfrac{{h'\left( x \right)}}{{h\left( x \right)}} = \dfrac{{{e^{{{\sin }^{ - 1}}x}} \times \dfrac{1}{{\sqrt {1 - {x^2}} }}}}{{{e^{{{\sin }^{ - 1}}x}}}}$
Cancelling similar terms, we get
$\Rightarrow \dfrac{{h'\left( x \right)}}{{h\left( x \right)}} = \dfrac{1}{{\sqrt {1 - {x^2}} }}$

So our answer is $\dfrac{1}{{\sqrt {1 - {x^2}} }}$.

Note:
We need to keep in mind to use different steps to form the composition function between two functions so that we may not get confused. Composition of two functions is applying one function to the result of another. We should remember which function is coming first and which is coming second otherwise the result will be different. The composite function is always associative and this is a property taken from composition of relation.