Question

Find the derivative of $y = \dfrac{{\ln 2x}}{{{e^{2x}} + 2}}$.

Answer 1

The given function is a fraction. So we can use quotient rules for finding derivatives. For that, we have to find the derivative of the numerator and denominator separately. Then substituting and simplifying we get the answer.

Formula used: Quotient rule for finding derivatives:
For a function $y = \dfrac{{f(x)}}{{g(x)}}$, $\dfrac{{dy}}{{dx}} = \dfrac{{g(x)f'(x) - f(x)g'(x)}}{{{{(g(x))}^2}}}$
Where $f'(x),g'(x)$ denotes the derivatives of the functions $f(x),g(x)$.
$\dfrac{{d(\ln x)}}{{dx}} = \dfrac{1}{x}$
$\dfrac{{d{e^x}}}{{dx}} = {e^x}$
Chain rule: To find the derivative of composite functions.
Let $y = f(g(x))$, $\dfrac{{dy}}{{dx}} = f'(x)g(x) \cdot g'(x)$

Complete step-by-step answer:
The given function is $y = \dfrac{{\ln 2x}}{{{e^{2x}} + 2}}$
We can see the function is the quotient of two functions $\ln 2x$ and ${e^{2x}} + 2$.
We need to find its derivative.
For a function, $y = \dfrac{{f(x)}}{{g(x)}}$, $\dfrac{{dy}}{{dx}} = \dfrac{{g(x)f'(x) - f(x)g'(x)}}{{{{(g(x))}^2}}}$ (quotient rule of derivatives)
Where $f'(x),g'(x)$ denotes the derivatives of the functions $f(x),g(x)$.
So here we have $f(x) = \ln 2x$ and $g(x) = {e^{2x}} + 2$.
For, $y = f(g(x))$, $\dfrac{{dy}}{{dx}} = f'(x)g(x) \cdot g'(x)$
So here we have,
$f'(x) = \dfrac{1}{{2x}} \times 2 = \dfrac{1}{x}$ and $g(x) = {e^{2x}} \times 2 = 2{e^{2x}}$
Since $\dfrac{{d(\ln x)}}{{dx}} = \dfrac{1}{x}$ and $\dfrac{{d{e^x}}}{{dx}} = {e^x}$
Substituting we have,
$\dfrac{{dy}}{{dx}} = \dfrac{{({e^{2x}} + 2) \times \dfrac{1}{x} - (\ln 2x) \times 2{e^{2x}}}}{{{{({e^{2x}} + 2)}^2}}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{{e^{2x}}}}{x} + \dfrac{2}{x} - 2{e^{2x}}\ln 2x}}{{{{({e^{2x}} + 2)}^2}}}$

Therefore, derivative of the given function is $\dfrac{{\dfrac{{{e^{2x}}}}{x} + \dfrac{2}{x} - 2{e^{2x}}\ln 2x}}{{{{({e^{2x}} + 2)}^2}}}$.

Note: Like the quotient rule, we have other rules for finding the derivative.
(i) Product rule: To find the derivative of a product of two functions $f(x)$ and $g(x)$
Let $y = f(x)g(x)$, $\dfrac{{dy}}{{dx}} = f(x)g'(x) + g(x)f'(x)$
(ii) Chain rule: To find the derivative of composite functions.
Let $y = f(g(x))$, $\dfrac{{dy}}{{dx}} = f'(x)g(x) \cdot g'(x)$
$\ln x$ represents the natural logarithm or logarithm to the base of the mathematical constant $e$. If considered as a real valued function, it is the inverse function of the exponential function.