QUESTION IMAGE
Question
suppose f is differentiable on r. let f(x)=f(e^x) and g(x)=e^f(x). find an expression for each derivative.
(a) f(x)
- f(e^x)
- f(e^x)e^x
- f(e^x)xe^x
- f(e^x)e^x
- f(e^x)xe^x
(b) g(x)
- e^f(x)f(x)
- e^f(x)f(x)
- e^f(x)f(x)
- e^f(x)
- e^f(x)f(x)
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Step1: Apply chain - rule for $F(x)$
The chain - rule states that if $F(x)=f(g(x))$, then $F^{\prime}(x)=f^{\prime}(g(x))\cdot g^{\prime}(x)$. Here, $g(x) = e^{x}$, and $g^{\prime}(x)=e^{x}$. So, $F^{\prime}(x)=f^{\prime}(e^{x})\cdot e^{x}$.
Step2: Apply chain - rule for $G(x)$
If $G(x)=e^{f(x)}$, let $u = f(x)$. Then $G(x)=e^{u}$. By the chain - rule, $G^{\prime}(x)=\frac{dG}{du}\cdot\frac{du}{dx}$. Since $\frac{dG}{du}=e^{u}$ and $\frac{du}{dx}=f^{\prime}(x)$, substituting $u = f(x)$ back in, we get $G^{\prime}(x)=e^{f(x)}f^{\prime}(x)$.
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(a) $f^{\prime}(e^{x})e^{x}$
(b) $e^{f(x)}f^{\prime}(x)$