How do you differentiate $f (t) = {sin}^{2} (e^{{sin}^{2} t})$ $f(t)=sin^2(e^(sin^2t))$ using the chain rule?

Question

1834 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-02T00:29:08

$2 {sin e}^{{sin}^{2} t}$ $2sine^[sin^2t]$ ${cos e}^{{sin}^{2} t}$ $cose^[sin^2t]$

${sin}^{2} [e^{{sin}^{2} t}] = {[{sin e}^{{sin}^{2} t}]}^{2}$ $sin^2[e^[sin^2t]]= [sine^[Sin^2t]]^2$ so using the chain rule , we have ....

$2 [{sin e}^{{sin}^{2} t}]$ $2[sine^[sin^2t]]$ times the differential within the the brackets ie......

$cos [e^{{sin}^{2} t}]$ $cos[e^[sin^2t]]$ and so we get d/dt $= 2 {sin e}^{{sin}^{2} t}$ $=2sine^[sin^2t]$ ${cos e}^{{sin}^{2} t}$ $cose^[sin^2t]$ .

How do you differentiate f(t)=sin2(esin2t)f(t)=sin^2(e^(sin^2t)) using the chain rule?