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Question: How do you differentiate \(f\left( x \right) = 4{\sin ^2}x + 5{\cos ^2}x\) ?...

How do you differentiate f(x)=4sin⁡2x+5cos⁡2xf\left( x \right) = 4{\sin ^2}x + 5{\cos ^2}x ?

Explanation

Solution

In this question, we are given a trigonometric equation and we have been asked to differentiate the given equation. Both the terms will be following the same rule. First, keep the constant aside. And differentiate just the trigonometric ratio. You will have to use chain rule for this purpose. Differentiate the first term, first and then differentiate the second term in the similar way.

Formula used: 1) d(sin⁡x)dx=cos⁡x\dfrac{{d\left( {\sin x} \right)}}{{dx}} = \cos x
2) d(cos⁡x)dx=−sin⁡x\dfrac{{d\left( {\cos x} \right)}}{{dx}} = - \sin x

Complete step-by-step solution:
We are given a trigonometric equation. Let us start differentiating it.
⇒f(x)=4sin⁡2x+5cos⁡2x\Rightarrow f\left( x \right) = 4{\sin ^2}x + 5{\cos ^2}x
Differentiate both the sides with respect to xx.
⇒f′(x)=4×2sin⁡x×d(sin⁡x)dx+5×2cos⁡x×d(cos⁡x)dx\Rightarrow f'\left( x \right) = 4 \times 2\sin x \times \dfrac{{d\left( {\sin x} \right)}}{{dx}} + 5 \times 2\cos x \times \dfrac{{d\left( {\cos x} \right)}}{{dx}}
Since we used chain rule to differentiate the trigonometric ratios, now we will do the second part of the chain rule.
⇒f′(x)=4×2sin⁡xcos⁡x+5×2cos⁡x×(−sin⁡x)\Rightarrow f'\left( x \right) = 4 \times 2\sin x\cos x + 5 \times 2\cos x \times \left( { - \sin x} \right)
Notice that we did not multiply the constants. This is because in the next step, we will be using the trigonometric identity of sin⁡2x\sin 2x to simplify the equation. So, it is not required to simplify it in one step and expand it again in the other step.
Using the formula sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x in the above equation,
⇒f′(x)=4sin⁡2x−5sin⁡2x\Rightarrow f'\left( x \right) = 4\sin 2x - 5\sin 2x
On simplifying it further, we will get,
⇒f′(x)=−sin⁡2x\Rightarrow f'\left( x \right) = - \sin 2x

Therefore, ⇒f′(x)=−sin⁡2x \Rightarrow f'\left( x \right) = - \sin 2x is our final answer.

Note: Chain rule: What is chain rule? This rule is used to find the derivative of composite functions. In a composite function, there is an outer function and an inner function. Correct identification of these functions is very important. Let us understand chain rule with the help of an example.
For example: (6x−x2)76{\left( {6x - {x^2}} \right)^{76}} is a composite function, where x76{x^{76}} is an outer function and 6x−x26x - {x^2} is an inner function. How do we find its derivative? At first, we differentiate the outer function and assume the inner function to be x (if we are differentiating with respect to x). In the next step, we differentiate the inner function.
Hence, correct identification of outer and inner function correctly is very important.