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Math Help - Fundemental Therom of Calculus with Chain Rule

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    Fundemental Therom of Calculus with Chain Rule

    Find \frac{d}{dx}\int^{x^4}_1 sec t, dt

    Solution:

    Let u = x^4

    \frac{d}{dx}\int^{x^4}_1 sec t, dt = \frac{d}{dx}\int^u_1 sect, dt

    = \frac{d}{du}(\int^{u}_1 sec t,dt) \frac{du}{dx}

    = sec u \frac{du}{dx}

    = sec({x^4})*{4x^3}

    Maybe I'm just rusty on the chain rule, I thought I needed to take the derivative of sec and then the derivative of what's inside the brackets: -tanxsecx * the derivative of u or t or whatever it is that would be inside the brackets.

    Could someone give me a rough walk through of how the textbook came up with the solution that it did?

    Thanks,

    Graeme
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  2. #2
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    Quote Originally Posted by Glorzifen View Post
    Find \frac{d}{dx}\int^{x^4}_1 sec t, dt

    Solution:

    Let u = x^4

    \frac{d}{dx}\int^{x^4}_1 sec t, dt = \frac{d}{dx}\int^u_1 sect, dt

    = \frac{d}{du}(\int^{u}_1 sec t,dt) \frac{du}{dx}

    = sec u \frac{du}{dx}

    = sec({x^4})*{4x^3}

    Maybe I'm just rusty on the chain rule, I thought I needed to take the derivative of sec and then the derivative of what's inside the brackets: -tanxsecx * the derivative of u or t or whatever it is that would be inside the brackets.

    Could someone give me a rough walk through of how the textbook came up with the solution that it did?

    Thanks,

    Graeme
    The conclusion of the Fundemental solution of calculus is

    let g(x)=\int_{a}^{x}f(t)dt then

    \frac{d}{dx}g(x)=\frac{d}{dx} \int_{a}^{x}f(t)dt=f(x)

    Now they are just using the chain rule.

    i.e u=x^4

    Then

    g(u)=\int_{a}^{u}f(t)dt

    \frac{d}{dx}g(u)=g'(u)\frac{du}{dx}=4x^3f(u)
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