Question

If y = log(tanx), then find $\dfrac{dy}{dx}$ at $x=\dfrac{\pi }{4}$.

Answer 1

Hint: Use the chain rule of differentiation, i.e. $\dfrac{d\left( g\left( f\left( x \right) \right) \right)}{dx}=\dfrac{d\left( gof\left( x \right) \right)}{d\left( f\left( x \right) \right)}\dfrac{d\left( f\left( x \right) \right)}{dx}$. Let g(x) = logx and f(x) = tanx. Apply chain rule of differentiation and hence find the derivative of gof(x). Finally, substitute $x=\dfrac{\pi }{4}$ and evaluate the expression of the derivative and hence find the value of the derivative at $x=\dfrac{\pi }{4}$. Alternatively, use the fact that if $y=\log x$ then $x={{e}^{y}}$ and hence prove that ${{e}^{y}}=\tan x$. Now differentiate both sides using the method of implicit differentiation to find the value of $\dfrac{dy}{dx}$. Substitute $x=\dfrac{\pi }{4}$ and the value of y at $x=\dfrac{\pi }{4}$ and hence find the value of $\dfrac{dy}{dx}$ at $x=\dfrac{\pi }{4}$.

Complete step-by-step answer:

We have $y=\log \left( \tan \left( x \right) \right)$
Let f(x) = tanx and g(x) = log(x)
Hence, we have $y=gof\left( x \right)$
Now, we know that from the chain rule of differentiation, $\dfrac{d\left( g\left( f\left( x \right) \right) \right)}{dx}=\dfrac{d\left( gof\left( x \right) \right)}{d\left( f\left( x \right) \right)}\dfrac{d\left( f\left( x \right) \right)}{dx}$
Hence, we have
$\dfrac{dy}{dx}=\dfrac{d}{d\left( f\left( x \right) \right)}\left( gof\left( x \right) \right)\dfrac{d}{dx}f\left( x \right)$
We know that $\dfrac{d}{dx}\log x=\dfrac{1}{x}$
Hence, we have $\dfrac{d}{d\left( f\left( x \right) \right)}g\left( f\left( x \right) \right)=\dfrac{1}{\tan x}$
We know that $\dfrac{d}{dx}\tan x={{\sec }^{2}}x$
Hence, we have $\dfrac{d}{dx}f\left( x \right)={{\sec }^{2}}x$
Hence, we have
$\dfrac{dy}{dx}=\dfrac{1}{\tan x}{{\sec }^{2}}x$
Now at $x=\dfrac{\pi }{4}$, we have $\tan x=1$ and $\sec x=\sqrt{2}$
Hence, we have
${{\left. \dfrac{dy}{dx} \right|}_{x=\dfrac{\pi }{4}}}=\dfrac{1}{1}{{\left( \sqrt{2} \right)}^{2}}=2$
Hence, the value of $\dfrac{dy}{dx}$ at $x=\dfrac{\pi }{4}$ is 2

Note: Alternative Solution:
We have $y=\log \left( \tan x \right)$
We know that if $y=\log x$ then $x={{e}^{y}}$
Hence, we have
${{e}^{y}}=\tan x$
Differentiating both sides, we get
${{e}^{y}}\dfrac{dy}{dx}={{\sec }^{2}}x$
Hence, we have $\dfrac{dy}{dx}=\dfrac{{{\sec }^{2}}x}{{{e}^{y}}}$
When $x=\dfrac{\pi }{4}$, we have $y=\log \left( \tan \left( \dfrac{\pi }{4} \right) \right)=\log 1=0$
Hence, we have
${{\left. \dfrac{dy}{dx} \right|}_{x=\dfrac{\pi }{4}}}=\dfrac{{{\left( \sqrt{2} \right)}^{2}}}{{{e}^{0}}}=2$, which is the same as obtained above.