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What is the first derivative of sin^4(x^2+1)?

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portoruj | Student, Grade 10 | eNoter

Posted December 29, 2010 at 10:22 PM via web

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What is the first derivative of sin^4(x^2+1)?

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giorgiana1976 | College Teacher | Valedictorian

Posted December 29, 2010 at 10:26 PM (Answer #1)

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We'll use the chain rule to differentiate the given function:

f'(x) = {[sin(x^2 + 1)]^4}'

We'll differentiate applying the power rule first, then we'll differentiate the sine function and, in the end, we'll differentiate the expression x^2 + 1.

f'(x) = 4[sin(x^2 +1)]^3*[cos(x^2 +1)]*(2x)

f'(x) = 8x[sin(x^2 +1)]^3*[cos(x^2 +1)]

We can re-write [sin(x^2 +1)]^2 = 1 - [cos(x^2 +1)]^2

f'(x) = 8x[sin(x^2 +1)]*[cos(x^2 +1)]*{1 - [cos(x^2 +1)]^2}

f'(x) = 4x*sin2(x^2 +1)*{1 - [cos(x^2 +1)]^2}

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted December 29, 2010 at 11:05 PM (Answer #2)

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Let f(x) = sin^4 (x^2 +1)

We need to find f'(x).

We will use the chain rule to find the derivative.

Let f(x) = u^4  such that: u= sin(x^2 +1)

==> f'(x) = 4u^3 * (u')

Now we will calculate u'.

==> u= sin (x^2+1) ==> u' = 2x*cos (x^2+1)

==> f'(x) = 4u^3 *2x*cos(x^2+1)

Now we will substitute with u = sin(x^2+1)

==> f'(x) = 8x*sin(x^2+1)]^3*cos(x^2+1)

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