Question 248534
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Everything is right except for one sign in the numerator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f^{\prime}(x)\ =\ \frac{g^{\prime}(x)h(x)\ -\ g(x)h^{\prime}(x)}{\left[h(x)\right]^2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ g(x)\ =\ 2x^2\ -\ 10]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ g^{\prime}(x)\ =\ 4x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(x)\ =\ \left(x^2\ +\ 5\right)^2]


We need the chain rule for this one:


Let *[tex \LARGE u\ =\ x^2\ +\ 5], so *[tex \LARGE \frac{du}{dx}\ =\ 2x] and *[tex \LARGE \frac{dh}{du}\ =\ 2u\ =\ 2x^2\ + 10], hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h^{\prime}(x)\ =\ 4x^3\ +\ 20x]


Putting it all together:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f^{\prime}\ =\ \frac{4x\left(x^2\ -\ 5\right)^2\ -\ \left(2x^2\ -\ 10\right)\left(4x^3\ +\ 20x\right)}{\left[x^2\ +\ 5\right]^4}]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f^{\prime}\ =\ \frac{4x\left(x^4\ -\ 10x^2\ +\ 25\right)\ -\ \left(8x^5\ +\ 40x^3\ -\ 40x^3\ -\ 200x\right)}{\left[x^2\ +\ 5\right]^4}]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f^{\prime}\ =\ \frac{4x^5\ -\ 40x^3\ +\ 100x\ -\ 8x^5\ +\ 200x}{\left[x^2\ +\ 5\right]^4}]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f^{\prime}\ =\ \frac{-4x^5\ -\ 40x^3\ +\ 300x}{\left[x^2\ +\ 5\right]^4}]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f^{\prime}\ =\ \frac{4x\left(-x^4\ -\ 10x^2\ +\ 75\right)}{\left[x^2\ +\ 5\right]^4}]


Though I might have written it thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f^{\prime}\ =\ \frac{-4x\left(x^4\ +\ 10x^2\ -\ 75\right)}{\left[x^2\ +\ 5\right]^4}]


But that is just changing 'happy' to 'glad' actually.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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