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.