suppose f is differentiable on r. let f(x)=f(e^x) and g(x)=e^f(x). find an expression for each…

suppose f is differentiable on r. let f(x)=f(e^x) and g(x)=e^f(x). find an expression for each derivative.\n(a) f(x)\n - f(e^x)\n - f(e^x)e^x\n - f(e^x)xe^x\n - f(e^x)e^x\n - f(e^x)xe^x\n(b) g(x)\n - e^f(x)f(x)\n - e^f(x)f(x)\n - e^f(x)f(x)\n - e^f(x)\n - e^f(x)f(x)\nresources\nread it

suppose f is differentiable on r. let f(x)=f(e^x) and g(x)=e^f(x). find an expression for each derivative.\n(a) f(x)\n - f(e^x)\n - f(e^x)e^x\n - f(e^x)xe^x\n - f(e^x)e^x\n - f(e^x)xe^x\n(b) g(x)\n - e^f(x)f(x)\n - e^f(x)f(x)\n - e^f(x)f(x)\n - e^f(x)\n - e^f(x)f(x)\nresources\nread it

Answer

Explanation:

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 $G^{\prime}(x)=e^{f(x)}f^{\prime}(x)$.

Answer:

(a) $f^{\prime}(e^{x})e^{x}$ (b) $e^{f(x)}f^{\prime}(x)$