Find the first positive x value of x for which f(x)=sin(e^x) has an inflection point. Correctly take derivatives.
Correct. Since $\displaystyle 0<e^x$ for all real $\displaystyle x$, this leaves you to solve:
$\displaystyle f(x)=\cos(e^x)-e^x\sin(e^x)=0$
I recommend a numeric root-finding technique, such as Newton's method. Graph the function to determine a good initial guess.
Hello, camjenson!
Your second derivative is incorrect.
Find the first positive value of $\displaystyle x$ for which $\displaystyle f(x)\:=\:\sin(e^x)$ has an inflection point.
We have: .$\displaystyle f(x) \:=\:\sin(e^x)$
Then: .$\displaystyle f'(x) \:=\:e^x\cdot\cos(e^x)$
Hence: .$\displaystyle f''(x) \;=\; e^x\!\cdot\!\cos(e^x) + e^x\!\cdot\![-\sin(e^x)]\!\cdot\! e^x \;=\;e^x\big[\cos(e^x) - e^x\sin(e^x)\big]$
And we must solve: .$\displaystyle \begin{Bmatrix}e^x \:=\:0 & [1] \\ \cos(e^x) - e^x\sin(e^x) \:=\:0 & [2] \end{Bmatrix}$
Equation [1] has no solutions.
Equation [2] can be approximated.
Let $\displaystyle u = e^x$
We have: .$\displaystyle \cos u - u\sin u \:=\:0 \quad\Rightarrow\quad u\sin u \:=\:\cos u \quad\Rightarrow\quad \frac{\sin u}{\cos u} \:=\:\frac{1}{u}$
And we can approximate a root of: .$\displaystyle \tan u \:=\:\frac{1}{u}$
Then back-substitute: .$\displaystyle x \,=\,\ln(u)$