Using the chain rule to differentiate composite functions (exponential of an exponential) How would I differentiate the following two functions using the chain rule.

*

*$t=(e^{2y})^{e^{y^2}+1}$


*$t=e^{{(y^3-\ln(y)+1)}^5}$.
 A: $$t(y)=(e^{2y})^{e^{y^2}+1}=e^{(2y)(e^{y^2}+1)}=e^{f(y)}=g(f(y))\\
t'(y)=(g(f(y)))'=f'(y)g'(f(y))$$
since we have defined $g(x)$ as $e^x$, we get $g'(x)=e^x$, thus
$$\boxed{g'(f(y))=e^{f(y)}=e^{(2y)(e^{y^2}+1)}}$$
now, we need to differentiate $f(y)=(2y)(e^{y^2}+1)=g(y)h(y)$, with $g(y)=2y$ and $h(y)=e^{y^2}+1$.
$$f'(y)=g'(y)h(y)+g(y)h'(y)$$
calculating $g'(y)$ is easy; $\boxed{g'(y)=2}$.
as for $h'(y)$, then you could further look at $h(y)=e^{y^2}+1$ as a composite function $h(y)=e^{z(y)}+1=w(z(y))$, with $z(y)=y^2$ and $w(x)=e^x+1$.
$$h'(y)=(w(z(y)))'=z'(y)w'(z(y))$$
easily, we get $z'(y)=2y$ and $w'(x)=e^x$, thus $w'(z(y))=e^{z(y)}=e^{y^2}$.
$$\boxed{h'(y)=2ye^{y^2}}$$
let's put everything together:
$$\boxed{f'(y)=2(e^{y^2}+1)+2y(2ye^{y^2})}$$
and, finally:
$$t'(y)=f'(y)g'(f(y))=\left[2(e^{y^2}+1)+2y(2ye^{y^2})\right]e^{(2y)(e^{y^2}+1)}$$
the rationale for the second function is the same.
A: hint
For the first:
$$t=e^{e^{y^2+1}\ln(e^{2y})}$$
$$=e^{2ye^{y^2+1}}=e^{f(y)}$$
the derivative is
$$f'(y)e^{f(y)}$$
with
$$f(y)=2ye^{y^2+1}=2g(y)e^{h(y)}$$
and
$$f'(y)=2\Bigl(g'(y)+g(y)h'(y)\Bigr)e^{h(y)}$$
A: $t=e^{(y^3−log(y)+1)^5}$
Here $t=e^{f(y)}$ and so $t' = e^{f(y)}\cdot f'(y)$.
Above $f(y) = (y^3−log(y)+1)^5$ and so $f'(y) = 5(y^3−log(y)+1)^4\cdot(3y^2 -\frac{1}{y})$.
A: I don't know what book you are using, but the rule to differentiate $k^m$ is just a combination of the exponential rule and the power rule:
$$ d(k^m) = mk^{m-1}\,dk + \ln(k)\,k^m\,dm $$
So, for the first one, just set $k = e^{2y}$ and $m = e^{y^2+1}$.
This means that $dk = 2e^{2y}\,dy$ and $dm = 2y\,e^{y^2 + 1}\,dy$.
Putting those in gives:
$$ (e^{y^2+1})^{e^{2y} - 1} e^{2y}\frac{dy}{dt} + \ln(e^{2y}) (e^{2y})^{e^{y^2+1}}(2y\,e^{y^2 + 1})\frac{dy}{dt} $$
You can use the same process for the second one as well.
