Question

What is the derivative of$y = {\left( {\sin x} \right)^{2x}}$ ?

Answer 1

Hint: Here we find the derivative of a given function by using implicit differentiation, product rule, and also chain rule. Implicit differentiation is a way of differentiating the function in terms of both $x$and$y$.
Chain rule: In differential calculus, we use the chain rule when we have a composite function and the rule states that the derivative of the outside function with respect to the inside times is the derivative of the inside function.
Formula used:
Product rule : $(uv)' = uv' + vu'$

Complete answer:
Given $y = {\left( {\sin x} \right)^{2x}}$----------------(1)
We have to find $\dfrac{{dy}}{{dx}}$ .
Now, we shall take log on both sides of the equation(1) we get,
$\log y = \log {\left( {\sin x} \right)^{2x}}$ --------------(2)
Using the logarithmic property that is $\log {m^n} = n\log m$ in equation (2) we get,
$\log y = 2x\,.\log \left( {\sin x} \right)$ ------------(3)
Now we need to apply the implicit differentiation, chain rule, and also the product rule to find the $\dfrac{{dy}}{{dx}}$
We know that $\dfrac{{d\left( {\log x} \right)}}{{dx}} = \dfrac{1}{x}$ and $\dfrac{{d(kx)}}{{dx}} = k$ ,$\dfrac{{d(\sin x)}}{{dx}} = \cos x$
From this, we can differentiate equation (3) we get,
$\dfrac{1}{y}\dfrac{{dy}}{{dx}} = 2x.\dfrac{1}{{\sin x}}\left( {\cos x} \right) + \log \left( {\sin x} \right).2$
Here we used implicit differentiation to the left-hand side and product rule, chain rule on the left-hand side.
We want only $\dfrac{{dy}}{{dx}}$to hold this term on the left-hand side and move the other term into the right-hand side we get,
$\dfrac{{dy}}{{dx}} = y\left( {2x.\dfrac{{\cos x}}{{\sin x}} + \log \left( {\sin x} \right).2} \right)$
Now, we shall take the common terms outside the bracket, we get,
$\dfrac{{dy}}{{dx}} = 2y\left( {x.\dfrac{{\cos x}}{{\sin x}} + \log \left( {\sin x} \right)} \right)$
We know that $\cot x = \dfrac{{\cos x}}{{\sin x}}$ substitute this in above, we get
$\dfrac{{dy}}{{dx}} = 2y\left( {x.\cot x + \log \left( {\sin x} \right)} \right)$
This equation must consist of all the variables are in terms of $x$so we can substitute for the term $y$ in the above equation that is $y = {\left( {\sin x} \right)^{2x}}$then we get,
$\dfrac{{dy}}{{dx}} = 2{\left( {\sin x} \right)^{2x}}\left( {x.\cot x + \log \left( {\sin x} \right)} \right)$
This is the final answer and it is the derivative of $y = {\left( {\sin x} \right)^{2x}}$

Note: To use implicit differentiation that you can treat the variable $y$ the same as an $x$ and when you differentiate it you multiply be $\dfrac{{dy}}{{dx}}$. Also, we need to apply some differential formulas according to the problem. Here we have applied the logarithmic property.