What is the derivative of f(t)=sin^2(e^{sin^2(t)}) ?

The derivative of f(t)=sin^2(e^{sin^2(t)}) is e^{sin^2(t)}sin(2e^{sin^2(t)})sin(2t)

What is the first derivative of f(t)=sin^2(e^{sin^2(t)}) ?

The first derivative of f(t)=sin^2(e^{sin^2(t)}) is e^{sin^2(t)}sin(2e^{sin^2(t)})sin(2t)

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derivative of

f (t) = sin^{2} (e^{s i n^{2} (t)})

e^{s i n^{2} (t)} sin (2 e^{s i n^{2} (t)}) sin (2 t)

Solution steps

\frac{d}{d t} (sin^{2} (e^{s i n^{2} (t)}))

Apply the chain rule:

2 sin (e^{s i n^{2} (t)}) \frac{d}{d t} (sin (e^{s i n^{2} (t)}))

= 2 sin (e^{s i n^{2} (t)}) \frac{d}{d t} (sin (e^{s i n^{2} (t)}))

\frac{d}{d t} (sin (e^{s i n^{2} (t)})) = cos (e^{s i n^{2} (t)}) e^{s i n^{2} (t)} sin (2 t)

= 2 sin (e^{s i n^{2} (t)}) cos (e^{s i n^{2} (t)}) e^{s i n^{2} (t)} sin (2 t)

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

= e^{s i n^{2} (t)} sin (2 e^{s i n^{2} (t)}) sin (2 t)

\frac{2}{x}, at x = \frac{1}{4}

y = \frac{- 2 x}{x ^{2} + 1}, (1, - 1)

x \to 5 lim (3 x^{2} - 2 x - 7)

f (x) = 2 x^{2} + 5 x - 3, (1, 4)

\frac{d}{d x} (A x e^{- x} cos (2 x) + B x e^{- x} sin (2 x))

What is the derivative of f(t)=sin^2(e^{sin^2(t)}) ?
The derivative of f(t)=sin^2(e^{sin^2(t)}) is e^{sin^2(t)}sin(2e^{sin^2(t)})sin(2t)
What is the first derivative of f(t)=sin^2(e^{sin^2(t)}) ?
The first derivative of f(t)=sin^2(e^{sin^2(t)}) is e^{sin^2(t)}sin(2e^{sin^2(t)})sin(2t)