How to find the second derivative of y = e ^ SiNx?

How to find the second derivative of y = e ^ SiNx?

y'=e^(sin(x))*cos(x);
y''=e^(sin(x))*cos(x)^2 - e^(sin(x))*sin(x)

Find the derivative of y = e ^ SiNx ^ 2

solution
y=e^sinx ²
y’=(e^sinx ²)'
=e^sinx ²× (sinx ²)'
=2xcosx ² e^sinx ²

Let y = SiNx square of E, find the n-square of the second derivative y + n-square of Y (0)

Because y = e ^ SiNx
So, y '= e ^ SiNx · cosx
So, y '' = e ^ SiNx · cosx SiNx · e ^ SiNx
=(cosx-sinx)·e^sinx
So, (y '') ^ n + y ^ n
=(cosx-sinx)·e^(nsinx)+e^(nsinx)
=(cosx-sinx+1)·e^(nsinx)

Derivative y = 2 ^ SiNx

y=2^sinx
Y '= 2 ^ SiNx * LN2 * cosx y' = a ^ x * LNA from y = a ^ x
=ln2cosx*2^sinx

Y = 2 ^ SiNx derivative

y'=(2^sinx )'=2^sinx *ln2*(sinx)'=2ln2sinxcosx

sinx+sin2x+sin3x+.+sinnx=? Seek a sum A problem in summing a sequence of numbers

Let s = SiNx + sin2x + sin3x +... + sinnx
Multiply both sides by 2Sin (x / 2) (x ≠ 2K π, K ∈ z)
2Sin (x / 2) s = cos (x / 2) - cos [(2n + 1) x / 2] = 2Sin (NX / 2) sin [(n + 1) x / 2]
So s = sin (NX / 2) sin [(n + 1) x / 2] ÷ sin (x / 2)