Question

Find the second derivative of: - \[\sin 3x\cos 5x\].

Answer 1

Hint: In the above question first of all we will have to know about the second derivative. The second derivative of a function f is the derivative of the derivative of the function f. The second derivative is expressed as \[\dfrac{{{d}^{2}}f}{d{{x}^{2}}}\] or \[f''\]. Since, we have a product of two functions, we can use the product rule of differentiation to solve it further.

Complete step-by-step answer:
In the above question we will use the chain rule and product rule of differentiation which is shown as below: - If we have two functions \[f\left( x \right).g\left( x \right)\]

\[\Rightarrow \] Chain Rule: - \[\dfrac{df\left( g\left( x \right) \right)}{dx}=f'\left( g\left( x \right) \right).g'\left( x \right)\]

\[\Rightarrow \] Product Rule: - \[\dfrac{d}{dx}\left( f\left( x \right).g\left( x \right) \right)=f'\left( x \right)g\left( x \right)+g'\left( x \right)f\left( x \right)\]

Now, in the above question we have been given the function as \[\sin 3x\cos 5x\].

So, we can find the first derivative of the function using the above rules as follows: -

\[\begin{align}

  & \dfrac{d}{dx}\left( \sin 3x.\cos 5x \right)=\sin 3x.\left( -\sin 5x \right)\times 5+\cos

5x\times \left( \cos 3x \right)\times 3 \\

 & \dfrac{d}{dx}\left( \sin 3x.\cos 5x \right)=-5\sin 3x\sin 5x+3\cos 5x\cos 3x \\

\end{align}\]

Again, we will have to differentiate the first differentiation to get the second derivative of the

function which is shown as below: -

\[\dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( \sin 3x.\cos 5x \right)=\dfrac{d}{dx}\left( -5\sin 3x\sin 5x

\right)+\dfrac{d}{dx}\left( 3\cos 5x\cos 3x \right)\]

Here, we will use the chain rule and product rule:

\[\dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( \sin 3x.\cos 5x \right)=-5\sin 3x.\left( \cos 5x \right)\times

 5-5\sin 5x\left( \cos 3x \right)\times 3+3\cos 5x\left( -\sin 3x \right)\times 3+3\cos x\left(

 -5\sin 5x \right)\times 5\]

\[\dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( \sin 3x.\cos 5x \right)=-25\sin 3x\cos 5x-15\sin 5x\cos

3x-9\cos 5x\sin 3x-15\cos 3x.\sin 5x\]

Therefore, the second derivative of the function given in the question is, \[-25\sin 3x\cos

5x-15\sin 5x\cos 3x-9\cos 5x\sin 3x-15\cos 3x.\sin 5x\]

Note: Just be careful while doing calculation as there is a chance that you might make a mistake while using chain rule and you will get the incorrect answer. Also, remember the rule of differentiation like sum rule, difference rule and division rule. Remember the fact about the second derivative that it measures the concavity and convexity of the graph of any function.