Differientiation of y=e^sinx

Differentiation of y=e^(sinx) wrt to x.

My method: ln both sides and then differientiate - is that okay?

I.e.

ln y = sinx ln e = sinx

1/y (dy/dx) = cosx

dy/dx = (cosx)(e^sinx)

OR:

I put u = sin x so du/dx = cos x; then y = e^u so dy/du = e^u; then
dy/dx = (e^u) cos x = cos x e^sin x. Now sin x isn't a constant so is this
wrong?

Thanks in advance!

Reply 1

CalculusMan

10

y = e^f(x), dy/dx = f'(x) e^f(x)

So f(x) = sin x, f'(x) = cos x

so dy/dx = cos x e^sin x

Your second answer is correct.

Reply 2

CalculusMan

10

First is incorrect method as you would need to differentiate both sides, terms of x and y, and I presume this is C3? Differentiating both isn't covered until C4.

Reply 3

Eau

OP

15

CalculusMan

y = e^f(x), dy/dx = f'(x) e^f(x)

So f(x) = sin x, f'(x) = cos x

so dy/dx = cos x e^sin x

Your second answer is correct.

So when can't you use this kind of method (second method). Because I know you can't use it for:

y = x^x

so why can you use it on y = x^(sinx)

thanks

Reply 4

silent ninja

19

They both look correct to me. In the first one you've differentiated implicitly-- i dont know is that in C3?

You can use the second method for y=x^x but that doesnt help you in any way. If you let u=x you just get y=x^u (or u^u) which doesnt help because you need to natural log both sides in order to differentiatie, so you may as well do that from the start.

Reply 5

DDCA

3

wildfire

Differentiation of y=e^(sinx) wrt to x.

My method: ln both sides and then differientiate - is that okay?

I.e.

ln y = sinx ln e = sinx

1/y (dy/dx) = cosx

dy/dx = (cosx)(e^sinx)

OR:

I put u = sin x so du/dx = cos x; then y = e^u so dy/du = e^u; then
dy/dx = (e^u) cos x = cos x e^sin x. Now sin x isn't a constant so is this
wrong?

Thanks in advance!

Both methods are fine.

Reply 6

CalculusMan

10

Sorry wildfire I didn't read your first method correctly - they are both correct, but your first one is implict differentiation, which isn't in C3, however, you could still use it, but the 2nd method is alot easier!

When, y = e^x, dy/dx = e^x. This only works with 'e'

y = a^x, dy/dx = a^x ln a. So y = x^x, dy/dx = x^x ln x.

You can't differentiate y = x^x with C3 methods.

Reply 7

silent ninja

19

y = x^x
lny = xlnx
1/y.dy/dx = x.1/x + lnx
dy/dx = y(1+lnx)
dy/dx = x^x(1+lnx)

Reply 8

investor

nice1

Differientiation of y=e^sinx

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