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Thread: derivative question

  1. #1
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    derivative question

    im doing question

    f(x) = (x^2 - [e^(2x)] + [sin^3 (x)])^7, find f'(x)

    so far,
    f'(x)= 7(x^2 - [e^(2x)] + [sin^3 (x)])^6 * (x^2 - [e^(2x)] + [sin^3 (x)])'
    = 7(x^2 - [e^(2x)] + [sin^3 (x)])^6 *@@@@@@@@@@@@@@

    for @@@@@@@@...
    i know i have to derivate (x^2 - [e^(2x)] + [sin^3 (x)])'
    i was wondering if i just derivate like 2x - 2e^2x + 3(sinx)^2 (cosx) ?

    if so, do i have to derivate "x" from"
    (sinx)^3
    3(sinx)^2 (cosx) <--- from here, do i have to derivate "x"? like 3(sinx)^2 (cosx)(x)' ?
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  2. #2
    Member mathemagister's Avatar
    Joined
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    Quote Originally Posted by haebinpark View Post
    im doing question

    f(x) = (x^2 - [e^(2x)] + [sin^3 (x)])^7, find f'(x)

    so far,
    f'(x)= 7(x^2 - [e^(2x)] + [sin^3 (x)])^6 * (x^2 - [e^(2x)] + [sin^3 (x)])'
    = 7(x^2 - [e^(2x)] + [sin^3 (x)])^6 *@@@@@@@@@@@@@@

    for @@@@@@@@...
    i know i have to derivate (x^2 - [e^(2x)] + [sin^3 (x)])'
    i was wondering if i just derivate like 2x - 2e^2x + 3(sinx)^2 (cosx) ?

    if so, do i have to derivate "x" from"
    (sinx)^3
    3(sinx)^2 (cosx) <--- from here, do i have to derivate "x"? like 3(sinx)^2 (cosx)(x)' ?
    I explained to you about the derivative of the 'x' in the other thread. You understand it, right?

    $\displaystyle f(x) = (x^2 - e^{2x} + sin^3x)^7$
    $\displaystyle f'(x) = 7(x^2 - e^{2x} + sin^3x)^6 \times (2x - e^{2x} \times 2 + 3sin^2x \times cosx)$

    I've pointed out the places where I've multiplied something because of the chain rule with $\displaystyle \times$. Do you see how that works. If you need anymore help, feel free to ask.

    Hope I helped you
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