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Find second and third derivative of \[e^{2x^{2}}\]
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\[e^{2x^{2}}\]
Use the chain rule\[\frac{d}{dx}f(g(x))=f'(g(x)) \times g'(x)\]So for the first derivative:\[\frac{d}{dx}e^{2x^{2}}=e^{2x^{2}} \times 4x\]
I prefer this kind of chain rule\[\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}\]\[y = e^{2x^{2}}\]\[u = 2x^{2}\]\[\frac{du}{dx} = 4x\]\[y = e^{u}\]\[\frac{dy}{du} = e^{u}\]\[\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx} = (e^u)(4x) = 4xe^{2x^{2}}\]for the second derivative, apply the product rule \[(uv)' = u'v + uv'\]
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