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Question

Question: What is the derivative of \({\sin ^2}\left( {5x} \right)\)?...

What is the derivative of sin2(5x){\sin ^2}\left( {5x} \right)?

Explanation

Solution

In the given problem, we are required to differentiate sin2(5x){\sin ^2}\left( {5x} \right) with respect to x. The given function is a composite function, so we will have to apply the chain rule of differentiation in the process of differentiation. So, differentiation of sin2(5x){\sin ^2}\left( {5x} \right) with respect to x will be done layer by layer. Also the derivative of sinx\sin xwith respect to x must be remembered.

Complete step by step solution:
So, Derivative of sin2(5x){\sin ^2}\left( {5x} \right) with respect to xxcan be calculated as ddx(sin2(5x))\dfrac{d}{{dx}}\left( {{{\sin }^2}\left( {5x} \right)} \right) .
Now, ddx(sin2(5x))\dfrac{d}{{dx}}\left( {{{\sin }^2}\left( {5x} \right)} \right)
Now, Let us assume u=sin5xu = \sin 5x. So substituting sin5x\sin 5x as uu, we get,
== ddx(u2)\dfrac{d}{{dx}}\left( {{u^2}} \right)
Now, we know the power rule of differentiation as d(xn)dx=nxn1\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}. So, we get,
=2u(dudx)= 2u\left( {\dfrac{{du}}{{dx}}} \right)
Now, putting back uuas sin5x\sin 5x, we get,
=2sin5x(d(sin5x)dx)= 2\sin 5x\left( {\dfrac{{d\left( {\sin 5x} \right)}}{{dx}}} \right) because dudx=d(sin5x)dx\dfrac{{du}}{{dx}} = \dfrac{{d\left( {\sin 5x} \right)}}{{dx}}
Now, taking t=5xt = 5x. We get,
=2sin5x(d(sint)dx)= 2\sin 5x\left( {\dfrac{{d\left( {\sin t} \right)}}{{dx}}} \right)
We know that derivative of sine is equal to cosine. So, we get,
=2sin5x[cost×dtdx]= 2\sin 5x\left[ {\cos t \times \dfrac{{dt}}{{dx}}} \right]
Now, putting back the value of t in the equation, we get,
=2sin5x[cos(5x)×d(5x)dx]= 2\sin 5x\left[ {\cos \left( {5x} \right) \times \dfrac{{d\left( {5x} \right)}}{{dx}}} \right]
Using the power rule of differentiation again, we get,
=2sin5x[cos(5x)×5]= 2\sin 5x\left[ {\cos \left( {5x} \right) \times 5} \right]
Simplifying the expression, we get,
=10sin5xcos5x= 10\sin 5x\cos 5x
Now, we know the double angle formula for sine as sin2x=2sinxcosx\sin 2x = 2\sin x\cos x. Hence, we get,
=5sin(10x)= 5\sin \left( {10x} \right)
So, the derivative of sin2(5x){\sin ^2}\left( {5x} \right) with respect to xx is sin10x\sin 10x.

Note:
The chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer. Answer to the given problem can also be reported as 10sin5xcos5x10\sin 5x\cos 5x before applying the double angle formula for sine function, but it is better to provide the final answer in the condensed form.