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# How do you differentiate y=e^(x^2)?

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#### Explanation

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#### Explanation:

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6
Sep 17, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x {e}^{{x}^{2}}$

#### Explanation:

Chain Rule - In order to differentiate a function of a function, say $y , = f \left(g \left(x\right)\right)$, where we have to find $\frac{\mathrm{dy}}{\mathrm{dx}}$, we need to do (a) substitute $u = g \left(x\right)$, which gives us $y = f \left(u\right)$. Then we need to use a formula called Chain Rule, which states that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$. In fact if we have something like $y = f \left(g \left(h \left(x\right)\right)\right)$, we can have $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{df}} \times \frac{\mathrm{df}}{\mathrm{dg}} \times \frac{\mathrm{dg}}{\mathrm{dh}}$

Here we have $y = {e}^{u}$, where $u = {x}^{2}$

Hence, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$

= $\frac{d}{\mathrm{du}} {e}^{u} \times \frac{d}{\mathrm{dx}} \left({x}^{2}\right)$

= ${e}^{u} \times 2 x = 2 x {e}^{{x}^{2}}$

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