y=f(x^2)\\f'(x)=\sqrt{5x-1}\\ What is \frac{dy}{dx}?

Question

y=f(x^2)\\f'(x)=\sqrt{5x-1}\\

What is \frac{dy}{dx}?

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    0
    2021-09-10T18:45:12+00:00

    use chain rule

    remember that \frac{d}{dx} g(h(x))=g'(h(x))h'(x) (mixing Lagrange’s notation and Leibniz’s notation)

    \frac{dy}{dx}=

    \frac{d}{dx}y=

    \frac{d}{dx}f(x^2)=

    if we treat f(x) as g(x) and x^2 as h(x) then setup parts

    f'(x)=\sqrt{5x-1} (given) so f'(x^2)=\sqrt{5x^2-1}

    and h'(x)=\frac{d}{dx}x^2=2x

    \frac{d}{dx}y=

    (\sqrt{5x^2-1})(2x)=

    2x\sqrt{5x^2-1}

    answer:  \frac{dy}{dx}=2x\sqrt{5x^2-1}

    0
    2021-09-10T18:45:24+00:00

    Answer: \frac{dy}{dx}=\sqrt{20x^3-4x}

    Step by step:

    For better understanding I am using a new variable z instead of original x and then substitute x^2 for z. This makes the use of the chain rule clearer. Let me know if you have questions.

    y=f(z)\\f'(z)=\frac{dy}{dz}=\sqrt{5z-1}\\\frac{dy}{dx}=\frac{dy}{dz}\frac{dz}{dx}\\z=x^2\\y=f(x^2)\\\frac{dy}{dx}=f'(x^2)\frac{d(x^2)}{dx}\\\frac{dy}{dx}=\sqrt{5x^2-1}\cdot2x\\\frac{dy}{dx}=\sqrt{20x^3-4x}

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