Question

How do you find the derivative of ${\sin ^2}(2x + 3)$?

Answer 1

Hint: We will use the chain rule which states that $\dfrac{d}{{dx}}\left[ {f(g(x))} \right] = f'(g(x)).g'(x)$ for any two functions f (x) and g (x). Then, we will get the required answer.

Complete step-by-step answer:
We are given that we are required to find the derivative of ${\sin ^2}(2x + 3)$.
We will assume that $f(x) = {x^2}$ and g (x) = sin (2x + 3)
Now, we have $f(g(x)) = {\sin ^2}(2x + 3)$
Now, we will use the chain rule which states that $\dfrac{d}{{dx}}\left[ {f(g(x))} \right] = f'(g(x)).g'(x)$
Using the above mentioned rule and $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$, we will get:
$ \Rightarrow \dfrac{d}{{dx}}\left[ {{{\sin }^2}(2x + 3)} \right] = 2\sin (2x + 3).\dfrac{d}{{dx}}\left[ {\sin (2x + 3)} \right]$ ………(1)
Now, we also can assume that u(x) = sin x and v (x) = 2x + 3.
Now, we will get: u(v(x)) = sin (2x + 3)
Again using the chain rule in this, we will get: $\dfrac{d}{{dx}}[\sin (2x + 3)] = 2\cos (2x + 3)$ because we know that derivative of sin x is cos x and $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$.
Putting this in equation number 1, we will then obtain the following equation:-
$ \Rightarrow \dfrac{d}{{dx}}\left[ {{{\sin }^2}(2x + 3)} \right] = 2\sin (2x + 3) \times 2\cos (2x + 3)$
Now, if we simplify the right hand side, we will then obtain the following equation:-
$ \Rightarrow \dfrac{d}{{dx}}\left[ {{{\sin }^2}(2x + 3)} \right] = 4\sin (2x + 3)\cos (2x + 3)$ ……………(2)
Now, we already have a formula given by the following expression:-
$ \Rightarrow $sin (2y) = 2 sin y. cos y
Replacing y by 2x + 3, we will then obtain the following expression:-
$ \Rightarrow $sin (4x + 6) = 2 sin (2x + 3). cos (2x + 3)
Putting this in equation number 2, we will then obtain the following equation:-
$ \Rightarrow \dfrac{d}{{dx}}\left[ {{{\sin }^2}(2x + 3)} \right] = 2\sin (4x + 6)$

Hence, the required answer is 2 sin (4x + 6).

Note:
The students must note that they must commit to memory the following formulas and identities:-
Chain rule: $\dfrac{d}{{dx}}\left[ {f(g(x))} \right] = f'(g(x)).g'(x)$
$\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$
$\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x$
$\dfrac{d}{{dx}}\left( c \right) = 0$, where c is any constant
The students must note that when we took the derivative of 2x + 3, we received 2 because we use the fact that derivative of (2x + 3) is equal to derivative of 2x + derivative of 3, which can be terms as: $\dfrac{d}{{dx}}\left( {2x + 3} \right) = \dfrac{d}{{dx}}(2x) + \dfrac{d}{{dx}}(3)$
Now, we will write this as: $\dfrac{d}{{dx}}\left( {2x + 3} \right) = 2\dfrac{d}{{dx}}(x) + \dfrac{d}{{dx}}(3)$
Now, we will use the formulas mentioned in 2 and 4 to get that derivative of 2x + 3 is 2.