Differentiate g(x)=x^3 * cos^2 x.Differentiate g(x)=x^3 * cos^2 x.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to implicit differentiate the function with respect to x, using the rules of differentiation, such that:

`(dg(x))/(dx) = (d(x^3))/(dx)*cos^2 x + x^3*(d(cos^2 x))/(dx)`

`(dg(x))/(dx) = 3x^2*cos^2 x + x^3*2cos x*(d(cos x))/(dx)`

`(dg(x))/(dx) = 3(x*cos x)^2 + 2x^2*(x*cos x)*(-sin x)`

Factoring out `x*cos x` yields:

`(dg(x))/(dx) = (x*cos x)*(3(x*cos x) - 2x*(x*sin x))`

Factoring out x yields:

`(dg(x))/(dx) = (x^2*cos x)*(3(cos x) - 2(x*sin x))`

Hence, differentiating the given function, yields `(dg(x))/(dx) = (x^2*cos x)*(3(cos x) - 2(x*sin x).`

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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To calculate the first derivative of the given function, we'll use the product rule and the chain rule:

g'(x) = x^3 * (cos x)^2

We'll have 2 functions f and h:

(f*h)' = f'*h + f*h'

We'll put f = x^3 => f' = 3x^2

We'll put h = (cos x)^2 => h' = 2(cos x)*(cos x)'

h' = -2(sin x)*(cos x)

h' = - sin 2x

We'll substitute f,h,f',h' in the expression of (f*h)':

(f*h)' = 3x^2*(cos x)^2 - x^3*(sin 2x)

We'll factorize by x^2:

g'(x) = x^2[3*(cos x)^2 - x*(sin 2x)]