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DerivativeFind the first derivative of y=tan^4(x+1)^4

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sapon | Student, Undergraduate | eNoter

Posted June 1, 2011 at 3:58 PM via web

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Derivative

Find the first derivative of y=tan^4(x+1)^4

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted March 17, 2013 at 6:16 PM (Answer #3)

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You need use implicit differentiation, such that:

`(dy)/(dx) = (d(tan^4(x+1)^4))/(dx)`

`(dy)/(dx) = 4tan^3(x+1)^4*(d(tan(x+1)^4))/(dx)`

`(dy)/(dx) = 4tan^3(x+1)^4*(d(tan(x+1)^4))/(dx)`

`(dy)/(dx) = 4tan^3(x+1)^4*(1/(cos^2(x + 1)^4))*(d((x+1)^4))/(dx)`

`(dy)/(dx) = 4tan^3(x+1)^4*(1/(cos^2(x + 1)^4))*4(x+1)^3*(d(x+1))/(dx)`

`(dy)/(dx) = 4tan^3(x+1)^4*(1/(cos^2(x + 1)^4))*4(x+1)^3*1`

`(dy)/(dx) = 16(((x+1)tan(x+1)^4)^3)/(cos^2(x + 1)^4)`

Hence, evaluating the derivative of the given function, yields `(dy)/(dx) = 16(((x+1)tan(x+1)^4)^3)/(cos^2(x + 1)^4).`

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giorgiana1976 | College Teacher | Valedictorian

Posted June 2, 2011 at 1:30 AM (Answer #2)

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

We'll use the chain rule to differentiate the function.

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 replace 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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