# differentiate functiony=(x^7+3x^4+cosx)^2

Asked on by istetz

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have the function y=(x^7+3x^4+cosx)^2

To find the derivative use the chain rule

y = [f(x)]^2

=> y' = f'(x)*2*f(x)

For y=(x^7+3x^4+cosx)^2

The derivative y' = 2*(x^7+3x^4+cosx)*(7x^6 + 12x^3 - sin x)

The required derivative is y' = 2*(x^7+3x^4+cosx)*(7x^6 + 12x^3 - sin x)

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To use the chain rule, we'll specify first that f(x) is the result of composition of 2 functions.

u(x) = x^7 + 3x^4 + cos x and v(u) = u^2

f(x) = (vou)(x) = v(u(x)) = v(x^7 + 3x^4 + cos x) = (x^7 + 3x^4 + cos x)^2

We'll differentiate f(x) and we'll get:

f'(x) = v'(u(x))*u'(x)

First, we'll differentiate v with respect to u:

v'(u) = 2u^(2-1) = 2u

Second, we'll differentiate u with respect to x:

u'(x) = (x^7 + 3x^4 + cos x)' =7x^6 + 12x^3 - sin x

f'(x) = 2u*(7x^6 + 12x^3 - sin x)

We'll substitute u and we'll get:

The derivative of f(x) is: f'(x) = 2(x^7 + 3x^4 + cos x)*(7x^6 + 12x^3 - sin x)

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