I am working on a deeply nested function. I should find the $\frac{df}{dx}$ considering the following functions:
$a = x^2,\ b = \exp(a),\ c = a + b,\ d = \log{c},\ e = \sin{c},\ f = d + e$
My approach so far has been has been to first calculate $\frac{df}{dc}$, then $\frac{dc}{dx}$ and multiply them together. Then I substitute the appropriate values for $x$ in my answer. However I can't seem to arrive at the right answer.
I believe the problem is mostly how I am picking this problem apart. What would be a suggested approach to solve a deeply nested function like this?
Thanks so much!
These are not deeply nested functions. $$f(x)=\log{(x^2+e^{x^2})}+\sin{(x^2+e^{x^2})}$$ $$f'(x)=\frac{2x+2xe^{x^2}}{x^2+e^{x^2}}+(2x+2xe^{x^2})\cos{(x^2+e^{x^2})}$$ Assuming that $\log{(x)}$ is the natural logarithm.
One can also use the approach mentioned. $$f(x)=\log{(c(x))}+\sin{(c(x))}$$ $$f'(x)=\left(\frac{1}{c(x)}+\cos{(c(x))}\right)c'(x)$$ $$c(x)=a(x)+b(x)=a(x)+\exp{(a(x))}$$ $$c'(x)=a'(x)+a'(x)\exp{(a(x))}$$ $$a(x)=x^2$$ $$a'(x)=2x$$ $$\implies c'(x)=2x+2x\exp{(x^2)}$$ $$\implies f'(x)=\left(\frac{1}{x^2+\exp{(x^2)}}+\cos{(x^2+\exp{(x^2)})}\right)(2x+2x\exp{(x^2)})$$