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What is the derivative of f(x) = 3(x^3)* cosx

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samerrima | Student, Grade 9 | (Level 1) Honors

Posted November 14, 2010 at 6:19 AM via web

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What is the derivative of f(x) = 3(x^3)* cosx

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

Posted November 14, 2010 at 6:23 AM (Answer #1)

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f(x) = 3x^3 * cosx

First we will treat the funtion as a produt of two functions of x.

Let us assume that:

f(x) = u(x) *v(x)  such tht:

u(x)= 3x^3     ===>     u'(x) = 9x^2

v(x) = cosx    ===>       v'(x) = - sinx

Now we know that th product rule is:

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

        = 9x^2 *cosx + 3x^3 * -sinx

          = (9x^2)cosx - (3x^3)sinx

==> f'(x) = (9x^2)cosx - (3x^3)*sinx

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neela | High School Teacher | (Level 3) Valedictorian

Posted November 14, 2010 at 12:49 PM (Answer #2)

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To find the derivative of f(x) = 3x^3cosx.

We know that  d/dx{f(x)g(x) } = d/dxf(x)}g(x)+ f(x){d/dx g(x)}.

Therefore  d/dx {(3x^3)(cosx)} = {d/dx(3x^3)}cosx +(3x^3){d/dxcosx).

d/dx{((3x^3)(cosx)} = {3*3x^(3-1)}cosx + (3x^3)(-sinx), as d/dx (kx^n+ const) = knx^(n-1) and d/dx{(cosx) +a const) = -sinx.

 Therefore d/dx {(3x^3(cosx)} = 9x^2cosx -3x^3sinx.

d/dx {(3x^3)(cosx)} = 3x^2 {3xcosx-sinx}.

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Wiggin42 | Student, Grade 11 | (Level 2) Valedictorian

Posted April 26, 2014 at 1:16 AM (Answer #3)

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f(x) = 3(x^3)* cosx

This requires the product rule. 

f'(x)= (9x^2)cosx - 3sin(x)x^3

Take the derivative of the first function and multiply it by the second function. Take the derivative of the second function and multiply it by the first function. Add these two equations together to get the derivative. 

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