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Thread: Derivate of an Integral problem

  1. #1
    kym
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    Derivate of an Integral problem


    Why is D the correct option here? Shouldn't we just replace t^2 with (2x)^2? Why that 2 appears behind the sin?
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    Re: Derivate of an Integral problem

    $\displaystyle \dfrac{d}{dx} \bigg[\int_a^u f(t) \, dt \bigg] = f(u) \cdot \color{red}{\dfrac{du}{dx}}$

    $\displaystyle \dfrac{d}{dx} \bigg[\int_0^{2x} \sin(t^2) \, dt \bigg] = \sin[(2x)^2] \cdot \color{red}{2}$
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    Re: Derivate of an Integral problem

    Leibniz's rule: $\displaystyle \frac{d}{dx} \left(\int_{\alpha(x)}^{\beta(x)} f(x, t)dt\right)$$\displaystyle = f(x, \beta(x))\frac{d\beta(x)}{dx}- f(x, \alpha(x))\frac{d\alpha(x)}{dx}+$$\displaystyle \int_{\alpha(x)}^{\beta(x)}\frac{\partial f(x,t)}{\partial x} dt$.

    Here $\displaystyle \alpha(x)= 0$ so $\displaystyle \frac{d\alpha}{dx}= 0$, $\displaystyle \beta(x)= 2x$ so $\displaystyle \frac{d\beta}{dx}= 2$, and $\displaystyle f(x, t)= sin(t^2)$ so $\displaystyle \frac{\partial f(x, t)}{\partial x}= 0$

    $\displaystyle \frac{d}{dx} \left(\int_{\alpha(x)}^{\beta(x)} f(x, t)dt\right)= (sin((2x)^2)(2)= 2sin(4x^2)$
    Last edited by HallsofIvy; Nov 26th 2017 at 10:16 AM.
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