Question

find the derivative of f(w)=(8w^2 + 2)e^{w^2}
f(w)=

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Let $u = 8w^{2}+2$ and $v = e^{w^{2}}$.

Step2: Find $u'$

Differentiate $u = 8w^{2}+2$ with respect to $w$. Using the power - rule $\frac{d}{dw}(aw^{n})=naw^{n - 1}$, we get $u'=\frac{d}{dw}(8w^{2}+2)=16w$.

Step3: Find $v'$

Differentiate $v = e^{w^{2}}$ with respect to $w$ using the chain - rule. Let $t = w^{2}$, then $v = e^{t}$. The chain - rule is $\frac{dv}{dw}=\frac{dv}{dt}\cdot\frac{dt}{dw}$. We know that $\frac{dv}{dt}=e^{t}$ and $\frac{dt}{dw}=2w$. So $v'=e^{w^{2}}\cdot2w = 2we^{w^{2}}$.

Step4: Calculate $f'(w)$

Using the product - rule $f'(w)=u'v+uv'$, we substitute $u = 8w^{2}+2$, $u' = 16w$, $v = e^{w^{2}}$, and $v' = 2we^{w^{2}}$:

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Answer:

$(16w^{3}+20w)e^{w^{2}}$