Differentiate the Function: $f(t)= \sin^2(e^{\sin^2t})$ $f(t)= \sin^2(e^{\sin^2t})$
$f'(t)=2\cos(e^{\sin^2t})\cdot e^{\sin^2t}\cdot \cos^2t\cdot (-2\sin t)$
Is my answer correct?
 A: Here are the steps
$$\frac{d}{dt}\left[\sin^2\left(e^{\sin^2t}\right)\right]$$
$$=2\sin\left(e^{\sin^2t}\right)\frac{d}{dt}\left[\sin\left(e^{\sin^2t}\right)\right]$$
$$=2\sin\left(e^{\sin^2t}\right)\cos\left(e^{\sin^2t}\right)\frac{d}{dt}\left[e^{\sin^2t}\right]$$
$$=\sin\left(2e^{\sin^2t}\right)\frac{d}{dt}\left[e^{\sin^2t}\right]$$
$$=\sin\left(2e^{\sin^2t}\right)e^{\sin^2t}\frac{d}{dt}\left[\sin^2t\right]$$
$$=\sin\left(2e^{\sin^2t}\right)e^{\sin^2t}2\sin t\frac{d}{dt}\left[\sin t\right]$$
$$=\sin\left(2e^{\sin^2t}\right)e^{\sin^2t}2\sin t\cos t$$
$$=\sin\left(2e^{\sin^2t}\right)\sin(2t)e^{\sin^2t}$$
A: Your solution needs work. In particular, your use of the equal sign is misleading. Literally, $\sin^2(e^{\sin^2(t)}) \neq 2\cos(e^{\sin^2(t)}) \cdot e^{\sin^2(t)}\cos^2(t) \cdot (-2\sin(t))$.
I think you meant to write: $$f(t) = \sin^2(e^{\sin^2(t)})$$ and $$f'(t)=2\cos(e^{\sin^2(t)}) \cdot e^{\sin^2(t)}\cos^2(t) \cdot (-2\sin(t)).$$
This is something that many students do, and when I see it, I mark it as incorrect. It's important to work out good language skills in mathematics.
In anycase, this is the composition of several functions, and as such we need to use the chain rule:
$$f'(t) = \left( 2\sin(e^{\sin^2 t}) \right) \cdot \cos\left( e^{\sin^2(t)}\right) \cdot e^{\sin^2(t)} \cdot 2 \sin(t) \cdot \cos(t).$$
Let's rewrite our function as a composition of several other functions:
Let $g(t) = t^2$, $h(t) = \sin(t)$, $m(t) = e^t$, $n(t) = t^2$ and $p(t) = \sin(t)$, then $$f(t) = (g\circ h \circ m \circ n \circ p)(t) = g( h( m( n( p(t) ) ) ) )$$
Thus the derivative takes the form:
$$f'(t) = g'(h(m(n(p(t))))) \cdot h'(m(n(p(t))) \cdot m'(n(p(t)) \cdot n'(p(t)) \cdot p'(t).$$
A: $$f^{ \prime  }\left( t \right) =\left( \sin ^{ 2 }{ \left( { e }^{ \sin ^{ 2 }{ t }  } \right)  }  \right) ^{ \prime  }=2\sin { \left( { e }^{ \sin ^{ 2 }{ t }  } \right)  } \cos { \left( { e }^{ \sin ^{ 2 }{ t }  } \right)  } { e }^{ \sin ^{ 2 }{ t }  }2\sin { x } \cos { x } ={ e }^{ \sin ^{ 2 }{ t }  }\sin { 2\left( { e }^{ \sin ^{ 2 }{ t }  } \right)  } \sin { 2x } $$
