What is the first derivative of y=tan^4(x+1)^4 ?

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neela | High School Teacher | (Level 3) Valedictorian

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To find the first derivative of tan^4x(x+1)^4.

Let f(x) ={ tan(x+1)^4}^4.

We know d/dx(tanx )= sec^2x.

We know d/dx{f(x)}^n = n {f(x)}^(n-1)*f'(x).

We know d/dx (u(v(x)) = (d/dv){u(v(x)}{dv(x)/dx}.

Therefore d/dx{tan(x+1)^4}^4 = d/dv(x(tanv)^4 * dv/dx, where v = (x+1)^4.

d/dx{tan(x+1)^4}^4 = {4(tanv)^3sec^2v} dv/dx.

d/dx{tan(x+1)^4}^4 = {4[tan(x+1)^4]^3sec^2 (x+1)^4} d/dx(x+1)^4.

d/dx(tan(x+1)^4}^4 = {4[tan(x+1)^4]^3sec^2(x+1)^4}{4(x+1)^3

Therefore d/dx{tan(x+1)^4}^4 = 16(x+1)^3{[tan(x+1)^4]^3secx^2(x+1)^4}.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll differentiate a composed function:

f(x) = tan^4(x+1)^4

First, we'll differentiate the power function. We'll put

tan v(x)  = u(v(x))

(u^4)' = 4u^3

{[tan v(x)]^4}' = 4 [tan v(x)]^3*v'(x)

We'll put v(x) = (x+1)^4

v'(x) = 4(x+1)^3*(x+1)'

v'(x) = 4(x+1)^3

f'(x) = 4 [tan v(x)]^3*v'(x)

We'll substitute v(x) and v'(x) and we'll get:

f'(x) = 4 [tan(x+1)^4]^3*[4(x+1)^3]

f'(x) = 16[tan(x+1)^4]^3*[(x+1)^3]

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