what is the derivative of e^sinx???

Question

what is the derivative of tanx(e^sinx^2x)

Answer 1

It's not totally clear to me what you've typed; I'm going to assume it's

tan(x) * e^(sin^2(x))

first use product rule, then chain rule a bunch of times

d/dx [tan(x) * e^(sin^2(x))]
= tan(x) * d/dx [e^(sin^2(x))] + d/dx [tanx] * e^(sin^2(x))
= tan(x) * 2 * sinx * cosx *e^(sin^2(x)) + sec^2(x) * e^(sin^2(x))

Answer 2

derivative of e^sinx= (e^sinx)* (d/dx)(sinx) = (e^sinx)*(cosx)
derivative of tanx(e^sinx^2x)= tanx(e^sinx^2x)*(sinx^2x)log(sinx)*2+sec^2x(e^sinx^2x)

Answer 3

dy/dx = ((secx)^2)*(e^sinx^2x) + (tanx)*(x^2x)*(2lnx + 2)*cos(x^2x)*(e^sinx^2x)

Its a bit horrible maybe i slightly misunderstood exactly what you typed

Answer 4

1)y´= e^sinx * cos x
I suppose it is the product of tan(x*e^sin^2x)
y´= 1/cos^2(x*e^sin^2x) * (e^sin^2x + xe^sin^2x*2sin x* cosx)

Answer 5

e^sinx * cosx

Answer 6

If u r asking for y=e^sinx.
take ln y= sinx
=> (1/y)(dy/dx)=cos x
=> y'=cos x* e^sinx

If u r asking for y=tanx*e^sinx^2x,
let z=e^sinx^2x
also, let t=x^2x
then z=e^sint
=> dz/dt=cos t*e^sint
dz/dx=dz/dt*dt/dx
Since t=x^2x
ln t= 2x ln x
(1/t)(dt/dx)=2+2ln x
dt/dx=(x^2x)(2+2ln x)
So, dz/dx=cos(x^2x)*e^sin(x^2x)*(x^2x)(2+2ln x)

Since y=tanx*e^sinx^2x,
dy/dx = (sec x)^2* (e^sinx^2x)+tan x* (cos(x^2x)*e^sin(x^2x)*(x^2x)(2+2ln x))