
What is the derivative of ${\sin ^2}\left( {5x} \right)$?
Answer
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Hint:In the given problem, we are required to differentiate ${\sin ^2}\left( {5x} \right)$ 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 ${\sin ^2}\left( {5x} \right)$ with respect to x will be done layer by layer. Also the derivative of $\sin x$with respect to x must be remembered.
Complete step by step solution:
So, Derivative of ${\sin ^2}\left( {5x} \right)$ with respect to $x$can be calculated as $\dfrac{d}{{dx}}\left( {{{\sin }^2}\left( {5x} \right)} \right)$ .
Now, $\dfrac{d}{{dx}}\left( {{{\sin }^2}\left( {5x} \right)} \right)$
Now, Let us assume $u = \sin 5x$. So substituting $\sin 5x$ as $u$, we get,
$ = $$\dfrac{d}{{dx}}\left( {{u^2}} \right)$
Now, we know the power rule of differentiation as $\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$. So, we get,
$ = 2u\left( {\dfrac{{du}}{{dx}}} \right)$
Now, putting back $u$as $\sin 5x$, we get,
$ = 2\sin 5x\left( {\dfrac{{d\left( {\sin 5x} \right)}}{{dx}}} \right)$ because \[\dfrac{{du}}{{dx}} = \dfrac{{d\left( {\sin 5x} \right)}}{{dx}}\]
Now, taking $t = 5x$. We get,
$ = 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,
$ = 2\sin 5x\left[ {\cos t \times \dfrac{{dt}}{{dx}}} \right]$
Now, putting back the value of t in the equation, we get,
$ = 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,
$ = 2\sin 5x\left[ {\cos \left( {5x} \right) \times 5} \right]$
Simplifying the expression, we get,
$ = 10\sin 5x\cos 5x$
Now, we know the double angle formula for sine as $\sin 2x = 2\sin x\cos x$. Hence, we get,
$ = 5\sin \left( {10x} \right)$
So, the derivative of ${\sin ^2}\left( {5x} \right)$ with respect to $x$ is \[\sin 10x\].
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 $10\sin 5x\cos 5x$ before applying the double angle formula for sine function, but it is better to provide the final answer in the condensed form.
Complete step by step solution:
So, Derivative of ${\sin ^2}\left( {5x} \right)$ with respect to $x$can be calculated as $\dfrac{d}{{dx}}\left( {{{\sin }^2}\left( {5x} \right)} \right)$ .
Now, $\dfrac{d}{{dx}}\left( {{{\sin }^2}\left( {5x} \right)} \right)$
Now, Let us assume $u = \sin 5x$. So substituting $\sin 5x$ as $u$, we get,
$ = $$\dfrac{d}{{dx}}\left( {{u^2}} \right)$
Now, we know the power rule of differentiation as $\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$. So, we get,
$ = 2u\left( {\dfrac{{du}}{{dx}}} \right)$
Now, putting back $u$as $\sin 5x$, we get,
$ = 2\sin 5x\left( {\dfrac{{d\left( {\sin 5x} \right)}}{{dx}}} \right)$ because \[\dfrac{{du}}{{dx}} = \dfrac{{d\left( {\sin 5x} \right)}}{{dx}}\]
Now, taking $t = 5x$. We get,
$ = 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,
$ = 2\sin 5x\left[ {\cos t \times \dfrac{{dt}}{{dx}}} \right]$
Now, putting back the value of t in the equation, we get,
$ = 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,
$ = 2\sin 5x\left[ {\cos \left( {5x} \right) \times 5} \right]$
Simplifying the expression, we get,
$ = 10\sin 5x\cos 5x$
Now, we know the double angle formula for sine as $\sin 2x = 2\sin x\cos x$. Hence, we get,
$ = 5\sin \left( {10x} \right)$
So, the derivative of ${\sin ^2}\left( {5x} \right)$ with respect to $x$ is \[\sin 10x\].
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 $10\sin 5x\cos 5x$ before applying the double angle formula for sine function, but it is better to provide the final answer in the condensed form.
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