Question

How do you find the derivative of $y = \ln (\sin 2x)$ .

Answer 1

This is a combination of three functions $\ln$ , $\sin$ and $2\;x$ . So in this case we will use the chain rule of differentiation. Then we will differentiate the functions term by term. We may have to use the chain rule many times whenever there is a conjugate function.

Formula Used :
Chain rule of differentiation: The derivative of $f(g(x))$ is $f'(g(x)).g'(x)$ .
$\dfrac{d}{{dz}}\ln z = \dfrac{1}{z}$ .
$\dfrac{d}{{dz}}(\sin z) = \cos z$ .
$\ln u = v$
$\Rightarrow u = {e^v}$ .

Complete step by step answer:
We have;
$y = \ln (\sin 2x)$
Let, $f = \ln$
And $g = \sin$
And $h(x) = 2x$ .
$\therefore g(h(x)) = \sin 2x$ .
Then we can write;
$y = f(g(h(x)))$
At first, we will consider $g(h(x)) = j(x)$ .
$\therefore y = f(j(x))$
Differentiating both sides w.r.t. $x$ we will get;
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(f(j(x))$
Here we will apply the chain rule of differentiation and get;
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{df}}{{dx}}\dfrac{{dj}}{{dx}}$
Another form is;
$\Rightarrow \dfrac{{dy}}{{dx}} = f'(j(x)).j'(x)$
Now, we know that $\dfrac{d}{{dz}}\ln z = \dfrac{1}{z}$ . By applying this we will get;
$f'(j(x)) = \dfrac{1}{{\sin 2x}}$
Now, similarly by applying the chain rule of differentiation on $j(x)$ we will get;
$j'(x) = g'(h(x))h'(x)$
We know that $\dfrac{d}{{dz}}(\sin z) = \cos z$
$\therefore g'(h(x)) = \cos 2x$
And $h'(x) = \dfrac{d}{{dx}}(2x)$
Differentiating we get;
$\Rightarrow h'(x) = 2$
$\therefore j'(x) = (\cos 2x) \times 2$
Simplifying we get;
$\Rightarrow j'(x) = 2\cos 2x$
Another form is;
$\therefore \dfrac{{dy}}{{dx}} = \dfrac{{2\cos 2x}}{{\sin 2x}}$
We know that $\dfrac{{\cos \theta }}{{\sin \theta }} = {\text{cot}}\theta$ .
Applying this we will get;
$\Rightarrow \dfrac{{dy}}{{dx}} = 2\cot 2x$.
$\therefore$ Differentiating $y = \ln (\sin 2x)$ we get $2{\text{cot2x}}$ .

Note: This type of conjugate function will be easily solved by the chain rule of differentiation. But students must be careful about the differentiation of different functions and don’t forget to mention w.r.t. what you are differentiating. Most of the time the variable is ignored by the students and then the error occurs. In the alternative method, be careful about differentiating ${e^y}$ for $x$ .
Alternative Method:
We can solve this problem in another way.
Given;
$y = \ln (\sin 2x)$
We know that;
$\ln u = v$
$\Rightarrow u = {e^v}$
Applying this we will obtain;
${e^y} = \sin 2x$
Now if we will differentiate both sides w.r.t. $x$ we will get;
$\Rightarrow \dfrac{d}{{dx}}({e^y}) = \dfrac{d}{{dx}}(\sin 2x)$
Now applying the chain rule of differentiation;
$\dfrac{d}{{dx}}({e^y}) = {e^y}\dfrac{{dy}}{{dx}}$
We have seen that $\dfrac{d}{{dx}}(\sin 2x) = 2\cos 2x$ ;
Putting these values we will obtain;
${e^y}\dfrac{{dy}}{{dx}} = 2\cos 2x$
Dividing both sides with ${e^y}$ ;
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2\cos 2x}}{{{e^y}}}$
We have seen ${e^y} = \sin 2x$ .
$\therefore \dfrac{{dy}}{{dx}} = \dfrac{{2\cos 2x}}{{\sin 2x}}$
After simplification we will get;
$\Rightarrow \dfrac{{dy}}{{dx}} = 2\cot 2x$