Question

If we have a function as $f\left( x \right)=\dfrac{{{a}^{x}}}{{{x}^{a}}}$ then $f'\left( a \right)=$.
(A). $\log a-1$
(B). $\log a-a$
(C). $a\log a-a$
(D). $a\log a+a$

Answer 1

Hint: Assume the given function as y. Now apply logarithm on both sides. Now you have a difference of two terms instead of division. The difference can be differentiated easily. Now apply differentiation with respect to x on both sides logarithms can be vanished by differentiation. Now multiply with y on both sides. Use basic differentiation properties to solve all differential terms. Now substitute y as assumed function. Now you have the term $\dfrac{dy}{dx}$ on the left hand side which is nothing but $f'\left( x \right)$. So, you get $f'\left( x \right)$ result directly, use these formulas while calculating.
$\dfrac{d}{dx}\left( {{a}^{x}} \right)={{a}^{x}}\ln a$ ; $\dfrac{d}{dx}\left( {{x}^{a}} \right)=a{{x}^{a-1}}$

Complete step-by-step solution -
Given function in question, for which we need the derivative is:
$\Rightarrow f\left( x \right)=\dfrac{{{a}^{x}}}{{{x}^{a}}}$ - (1)
Assume the $f\left( x \right)$ to be y, we get it as below:
$\Rightarrow y=f\left( x \right)$ - (2)
By differentiating on both sides with respect to x, we get it as:
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}f\left( x \right)$
By simplifying the right hand side, we get it as below:
$\Rightarrow \dfrac{dy}{dx}=f'\left( x \right)$ - (3)
By simplifying equation (2) in equation (1), we get an equation:
$\Rightarrow y=\dfrac{{{a}^{x}}}{{{x}^{a}}}$
Applying log on both sides of equation, we get it as:
$\Rightarrow \log y=\log {{a}^{x}}-\log {{x}^{a}}$
We know $\dfrac{d}{dx}\log x=\dfrac{1}{x}$. So, use this here after differentiating.
By differentiating on both sides with respect to x, we get as;
\Rightarrow $$$$\dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{1}{{{a}^{x}}}\dfrac{d{{a}^{x}}}{dx}-\dfrac{1}{{{x}^{a}}}\dfrac{d{{x}^{a}}}{dx}
By using the formula $\dfrac{d}{dx}{{a}^{x}}={{a}^{x}}\log a$, $\dfrac{d{{x}^{k}}}{dx}=k{{x}^{k-1}}$ we get it as:
$\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{1}{{{a}^{x}}}{{a}^{x}}\log a-\dfrac{a.{{x}^{a-1}}}{{{x}^{a}}}$
By multiplying with y on both sides and simplifying, we get it as:
$\Rightarrow \dfrac{dy}{dx}=\dfrac{{{a}^{x}}}{{{x}^{a}}}\left( \log a-\dfrac{a}{x} \right)$
By substituting x = a, in the above equation we get:
$\Rightarrow f'\left( a \right)=\dfrac{{{a}^{a}}}{{{a}^{a}}}\left( \log a-\dfrac{a}{a} \right)=\log a-1$
So, we get the value of $f'\left( a \right)$ as $\log a-1$.
Therefore option (a) is correct.

Note: Alternately you can use logarithm property $\log {{a}^{b}}=b\log a$ to make the solution simpler. You will get differentiation of y in a simple step in this alternate method. Whenever you see the division of complex terms always apply logarithm before differentiation as it makes the solution simpler. Carefully observe that at x = a, we get ${{x}^{a}}={{a}^{x}}$. So, they cancel each other.