Question #3a2d9

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Question #3a2d9

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Answer 1 · 2017-10-23T18:21:29

$\frac{d y}{d x} = \frac{2 x}{5} \cdot 5 {(\frac{x^{2} + 4}{5})}^{4} = 2 x {(\frac{x^{2} + 4}{5})}^{4}$ $dy/dx = (2x)/5 * 5((x^2+4)/5)^4 = 2x((x^2+4)/5)^4$

Explanation:

We can most easily do this via the Chain Rule. The Chain Rule states that, given a function composition $f (g (x)), \frac{d f}{d x} = \frac{d g}{d x} \cdot \frac{d f}{d g}$ $f(g(x)), (df)/dx = (dg)/dx *(df)/(dg)$ . Considering the function $e^{x^{2}}$ $e^(x^2)$ , we have $g (x) = x^{2}, f (h) = e^{g}$ $g(x) = x^2, f(h) = e^(g)$ . Then $\frac{d g}{d x} = 2 x, \frac{d f}{d g} = e^{g}$ $(dg)/dx = 2x, (df)/(dg) = e^(g)$ , giving us $\frac{d f}{d x} = 2 x \cdot e^{x^{2}}$ $(df)/dx = 2x*e^(x^2)$

For this problem, we would have $y (x) = f (g (x)), g (x) = \frac{x^{2} + 4}{5}, f (g) = g^{5}$ $y(x) = f(g(x)), g(x) = (x^2+4)/5, f(g) = g^5$ . Then $\frac{d g}{d x} = \frac{2 x}{5}, \frac{d f}{d g} = 5 g^{4}$ $(dg)/dx = (2x)/5, (df)/(dg) = 5g^4$ . Substituting $\frac{x^{2} + 4}{5}$ $(x^2+4)/5$ back in for g, we get...

$\frac{d y}{d x} = \frac{2 x}{5} \cdot 5 {(\frac{x^{2} + 4}{5})}^{4} = 2 x {(\frac{x^{2} + 4}{5})}^{4}$ $dy/dx = (2x)/5 * 5((x^2+4)/5)^4 = 2x((x^2+4)/5)^4$

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Question #3a2d9

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