Question

Differentiate ${5^x}$ with respect to ${\log _5}x$.

Answer 1

Here, we will take that $h = {5^x}$ and $g = {\log _5}x$. Then we will use that when $h$ is differentiated with respect to $g$, we have to calculate the value of $\dfrac{{dh}}{{dg}}$. After differentiating $h$ with respect to $x$ and $g$ with respect to $x$, we will divide them to find the required value.

Complete step by step solution:
Let us assume that $h = {5^x}$ and $g = {\log _5}x$.
We know that when $h$ is differentiated with respect to $g$, we have to calculate the value of $\dfrac{{dh}}{{dg}}$.
Differentiating the equation $h$ with respect to $x$, we get
$\Rightarrow \dfrac{{dh}}{{dx}} = \dfrac{d}{{dx}}\left( {{5^x}} \right)$
Using the property, $\dfrac{d}{{dx}}{a^x} = {a^x}\log a$ in the above equation, we get
$\Rightarrow \dfrac{{dh}}{{dx}} = {5^x}\log 5{\text{ ......eq.(1)}}$
Using the logarithm property, ${\log _a}b = \dfrac{{\log b}}{{\log a}}$ in the equation $g = {\log _5}x$, we get
$\Rightarrow g = \dfrac{{\log x}}{{\log 5}}$
Differentiating the equation $g$ with respect to $x$, we get
$\Rightarrow \dfrac{{dg}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{{\log x}}{{\log 5}}} \right)$
Using the property, $\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$ in the above equation, we get
$\Rightarrow \dfrac{{dg}}{{dx}} = \dfrac{1}{{x\log 5}}{\text{ ......eq.(2)}}$
Dividing the equation (1) by equation (2), we get

$$\Rightarrow \dfrac{{\dfrac{{dh}}{{dx}}}}{{\dfrac{{dg}}{{dx}}}} = \dfrac{{{5^x}\log 5}}{{\dfrac{1}{{x\log 5}}}} \\\ \Rightarrow \dfrac{{dh}}{{dx}} \times \dfrac{{dx}}{{dg}} = {5^x}\log 5 \times x\log 5 \\\ \Rightarrow \dfrac{{dh}}{{dg}} = x{5^x}{\left( {\log 5} \right)^2} \\\$$

Hence, when ${5^x}$ is differentiated with respect to ${\log _5}x$, we get $x{5^x}{\left( {\log 5} \right)^2}$.

Note:
You should be familiar with the basic properties of differentiation and logarithm functions, like $\dfrac{d}{{dx}}{a^x} = {a^x}\log a$ and ${\log _a}b = \dfrac{{\log b}}{{\log a}}$. Students get confused to find the derivative and end up computing with respect to $x$, which is wrong. A function can only be differentiated with respect to another function if and only if both the functions are dependent on the same variable. The key point is to use the differentiation properly to find the final answer.