Question

How do you differentiate $f(x) = 2\cos x + \sin 2x$?

Answer 1

Hint: Here, we find the first order of the derivative and by using the chain rule of the composite function. First of all we will apply the derivative of the given power in the function and then by using the chain rule will take the derivative of the function and then the derivative of the angle.

Complete step-by-step answer:
Take the given expression:
$f(x) = 2\cos x + \sin 2x$
Now apply derivatives on both the sides of the equation.
$\dfrac{d}{{dx}}f(x) = \dfrac{d}{{dx}}(2\cos x + \sin 2x)$
When there is a derivative outside the bracket, the derivative is applied to all the terms inside the bracket.
$f'(x) = \dfrac{d}{{dx}}(2\cos x) + \dfrac{d}{{dx}}(\sin 2x)$
Take the constant outside the first term on the right hand side of the equation.
$f'(x) = 2\dfrac{d}{{dx}}(\cos x) + \dfrac{d}{{dx}}(\sin 2x)$
Here we will directly place the value for the derivative for the first term and for the second term we will use the chain rule.
 \[f'(x) = 2( - \sin x) + (\cos 2x)\dfrac{d}{{dx}}(2x)\]
Simplify the above equation and find the derivative of the angle for the second term.
 \[f'(x) = 2( - \sin x) + 2(\cos 2x)\dfrac{d}{{dx}}(x)\]
 \[f'(x) = 2( - \sin x) + 2(\cos 2x)(1)\]
Simplify the above equation –
 \[f'(x) = - 2\sin x + 2\cos 2x\]
The above equation can be re-written as –
 \[f'(x) = 2\cos 2x - 2\sin x\]
This is the required answer.
So, the correct answer is “ \[f'(x) = 2\cos 2x - 2\sin x\] ”.

Note: Know the difference between the differentiation and the integration and apply formula accordingly. Differentiation can be represented as the rate of change of the function, whereas integration represents the sum of the function over the range. They are inverses of each other. Remember the standard formulas for the derivatives and the integrations for different functions.