Question

Differentiate the following functions w. r. t. x
${\left( {{a^{\sqrt x }}} \right)^{\sin x}}$

Answer 1

We split the given function into 2 and write it as one function raised to another. Then we can find the derivatives of both the functions separately. Then we can use the Generalized Power Rule of differentiation which is given by the equation $\left( {{f^g}} \right)' = {f^g}\left( {f'\dfrac{g}{f} + g'\log f} \right)$ . Then we can substitute for the functions and its derivatives. After the substitution and simplification, we will obtain the required solution.

Complete step by step solution:
Let $y = {\left( {{a^{\sqrt x }}} \right)^{\sin x}}$ . Then its derivative is given by,
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}{\left( {{a^{\sqrt x }}} \right)^{\sin x}}$
We know that by Generalized Power Rule of differentiation, $\left( {{f^g}} \right)' = {f^g}\left( {f'\dfrac{g}{f} + g'\log f} \right)$
Here $f\left( x \right) = {a^{\sqrt x }}$ and $g\left( x \right) = \sin x$
Now we can find its derivative of $f\left( x \right)$ .
$\Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}{a^{\sqrt x }}$
We know that $\dfrac{d}{{dx}}{a^x} = {a^x}\log a$ . And by applying chain rule, we get,
$\Rightarrow f'\left( x \right) = {a^{\sqrt x }}\log a\dfrac{d}{{dx}}\sqrt x$
We know that $\dfrac{d}{{dx}}\sqrt x = \dfrac{1}{{2\sqrt x }}$ . On substituting this, we get,
$\Rightarrow f'\left( x \right) = {a^{\sqrt x }}\log a \times \dfrac{1}{{2\sqrt x }}$
Now we can find its derivative of $g\left( x \right)$ .
$\Rightarrow g'\left( x \right) = \dfrac{d}{{dx}}\sin x$
We know that $\dfrac{d}{{dx}}\sin x = \cos x$ . On applying this condition, we get,
$\Rightarrow g'\left( x \right) = \cos x$
Now we can use the Generalized Power Rule of differentiation.
On substituting the functions, we get,
$\Rightarrow \left( {{{\left( {{a^{\sqrt x }}} \right)}^{\sin x}}} \right)' = {\left( {{a^{\sqrt x }}} \right)^{\sin x}}\left( {{a^{\sqrt x }}\log a \times \dfrac{1}{{2\sqrt x }} \times \dfrac{{\sin x}}{{{a^{\sqrt x }}}} + \cos x \times \log {a^{\sqrt x }}} \right)$
We know that $\log \left( {{a^b}} \right) = b \times \log a$
$\Rightarrow \left( {{{\left( {{a^{\sqrt x }}} \right)}^{\sin x}}} \right)' = {\left( {{a^{\sqrt x }}} \right)^{\sin x}}\left( {{a^{\sqrt x }}\log a \times \dfrac{1}{{2\sqrt x }} \times \dfrac{{\sin x}}{{{a^{\sqrt x }}}} + \cos x \times \sqrt x \times \log a} \right)$
Now we can cancel the common terms and take $\log a$ outside the bracket,
$\Rightarrow \left( {{{\left( {{a^{\sqrt x }}} \right)}^{\sin x}}} \right)' = \log a{\left( {{a^{\sqrt x }}} \right)^{\sin x}}\left( {\dfrac{{\sin x}}{{2\sqrt x }} + \sqrt x \cos x} \right)$

Thus, the derivative of ${\left( {{a^{\sqrt x }}} \right)^{\sin x}}$ is $\log a{\left( {{a^{\sqrt x }}} \right)^{\sin x}}\left( {\dfrac{{\sin x}}{{2\sqrt x }} + \sqrt x \cos x} \right)$

Note:
The main property used here is the Generalized Power Rule of differentiation which is given by the equation $\left( {{f^g}} \right)' = {f^g}\left( {f'\dfrac{g}{f} + g'\log f} \right)$ . It is used for finding the derivatives of functions raised to some other function. We can derive the power rule by taking the second function as a constant function. We must take care while splitting the given expression into 2 functions. The order of the functions is important. As there are many functions in the equation, we must take care while substituting. We must apply the chain rule to find the derivative of the 1st function.