Question

Differentiate cos^4 x with respect to x

Answer 1

Answer:

Let u = cos xdu/dx = -sin xy = u4dy/du = 4u3dy/dx = dy/du * du/dx = (4u3 ) * (-sin x) = 4 cos 3 x * -sin x = -4 cos 3 x sin x

Answer 2

Answer:

- 4 sinxcos³x

Step-by-step explanation:

=> $\frac{d \:cos^{4}x }{dx}$

=> $4\:cos^{4-1}x$.($\frac{d \:cosx}{dx}$)

=> 4cos³x.(- sinx)

=> - 4sinxcos³x

                                                                                                                 

METHODOLOGY :-

[this is completed via chain rule followed in differentiation ]

=> d cos x/ dx = -sin x

=> $\frac{d x^{n} }{dx}$ = $n.(x^{n-1} )$

chain rule = if there are 2 functions one followed by the other then we solve the outermost function first and multiply its derivative by the derivative of th inner function.

Eg :-

cos⁴x = (cos x)⁴

here are 2 functions :-

1. exponential function
2. cos function

Step1:

the outermost function is exponential so finding its derivative :-

=> [$\frac{d x^{n} }{dx}$ = $n.(x^{n-1} )$]

=> $\frac{d \:cos^{4}x }{dx}$ = 4cos³x

Step 2:

now solving the inner function that is cos function :-

=> $\frac{d \:cosx}{dx} = - sin x$

Step 3:

multiplying both the derivatives :-

=> (4 cos³x). (- sinx)

=> -4sinxcos³x [Ans]

Hope I helped :)