## Calculus: Early Transcendentals 8th Edition

$$\frac{df}{dt}=\sin2e^{\sin^2t}\times e^{\sin^2t}\times\sin2t$$
$$f(t)=\sin^2(e^{\sin^2t})$$ $$\frac{df}{dt}=\frac{d(\sin^2(e^{\sin^2t}))}{dt}$$ Let $u=\sin t$, then $f=\sin^2(e^{u^2})$. According to Chain Rule, $$\frac{df}{dt}=\frac{df}{du}\frac{du}{dt}$$ $$\frac{df}{dt}=\frac{d(\sin^2(e^{u^2}))}{dt}\frac{d(\sin t)}{dt}$$ Let $v=u^2$, then $f=\sin^2(e^v)$. According to Chain Rule, $$\frac{df}{dt}=\frac{df}{dv}\frac{dv}{du}\frac{du}{dt}$$ $$\frac{df}{dt}=\frac{d(\sin^2(e^v))}{dv}\frac{d(u^2)}{du}\frac{d(\sin t)}{dt}$$ Let $q=e^v$, then $f=\sin^2q$. According to Chain Rule, $$\frac{df}{dt}=\frac{df}{dq}\frac{dq}{dv}\frac{dv}{du}\frac{du}{dt}$$ $$\frac{df}{dt}=\frac{d(\sin^2q)}{dq}\frac{d(e^v)}{dv}\frac{d(u^2)}{du}\frac{d(\sin t)}{dt}$$ Let $r=\sin q$, then $f=r^2$. According to Chain Rule, $$\frac{df}{dt}=\frac{df}{dr}\frac{dr}{dq}\frac{dq}{dv}\frac{dv}{du}\frac{du}{dt}$$ $$\frac{df}{dt}=\frac{d(r^2)}{dr}\frac{d(\sin q)}{dq}\frac{d(e^v)}{dv}\frac{d(u^2)}{du}\frac{d(\sin t)}{dt}$$ $$\frac{df}{dt}=2r\times\cos q\times e^v\times2u\times\cos t$$ $$\frac{df}{dt}=4\sin q\times\cos e^v\times e^{u^2}\times\sin t\times\cos t$$ $$\frac{df}{dt}=4\sin(e^v)\times\cos e^{u^2}\times e^{\sin^2t}\times\sin t\times\cos t$$ $$\frac{df}{dt}=4\sin(e^{u^2})\times\cos e^{\sin^2t}\times e^{\sin^2t}\times\sin t\times\cos t$$ $$\frac{df}{dt}=4\sin(e^{\sin^2t})\times\cos e^{\sin^2t}\times e^{\sin^2t}\times\sin t\times\cos t$$ $$\frac{df}{dt}=\sin2e^{\sin^2t}\times e^{\sin^2t}\times\sin2t$$ (since $2\sin x\cos x=\sin2x$)