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As before, the solutions here require the use of the 'chain rule'. The process is the same regardless of whether the exterior exponent is negative or even a fraction.

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For example,     $y=5(2x^4-7)^{-2}$

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Let $u=2x^4-7$

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This means that: $y=5u^{-2}\\;\\;\\;\\therefore\\;\\;\\;\\frac{dy}{du}=(5\\times-2)u^{-2-1}-10u^{-3}$

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Also: $u=2x^4-7\\;\\;\\;\\therefore\\;\\;\\;\\frac{du}{dx}=(2\\times4)x^{4-1}-(7\\times0)x^{0-1}=8x^3-0=8x^3$

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According to the chain rule:

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$\\frac{dy}{dx}=\\frac{dy}{du}\\times\\frac{du}{dx}=-10u^{-3}\\times8x^3=-80x^3u^{-3}$

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Subsituting the original expression back in the place of $u$ gives:

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$\\frac{dy}{dx}=-80x^3(2x^4-7)^{-3}$    which may also be written as    $\\frac{dy}{dx}=-\\frac{80x^3}{(2x^4-7)^3}$

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Or else, with a fractional index,    $y=\\sqrt[4]{x^2+9}=(x^2+9)^{\\frac{1}{4}}$

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Let $u=x^2+9$

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This means that: $y=u^{\\frac{1}{4}}\\;\\;\\;\\therefore\\;\\;\\;\\frac{dy}{du}=(1\\times\\frac{1}{4})u^{\\frac{1}{4}-1}=\\frac{1}{4}u^{-\\frac{3}{4}}$

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Also: $u=x^2+9\\;\\;\\;\\therefore\\;\\;\\;\\frac{du}{dx}=2x$

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According to the chain rule:

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$\\frac{dy}{dx}=\\frac{dy}{du}\\times\\frac{du}{dx}=\\frac{1}{4}u^{-\\frac{3}{4}}\\times2x=\\frac{2}{4}xu^{-\\frac{3}{4}}=\\frac{1}{2}xu^{-\\frac{3}{4}}$

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Subsituting the original expression back in the place of $u$ gives:

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$\\frac{dy}{dx}=\\frac{1}{2}x(x^2+9)^{-\\frac{3}{4}}$    which may also be written as    $\\dfrac{dy}{dx}=\\dfrac{x}{2(x^2+9)^{\\frac{3}{4}}}$

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$\\simplify{({c[2]}x^2+{d[2]})^({np[2]})}$

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$\\simplify{sqrt({c[5]}x+{d[5]})}$

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Differentiate the following polynomials with respect to $x$.

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Give your answer in the form $ax^{b}(cx^d+e)^f$, where $f$ is a negative power.

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It is not necessary to include powers of $1$ or terms to the power $0$ in your answer.

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negative power

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power

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coefficient

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Using the chain rule with polynomials and negative powers.

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