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Differential calculus

I'm stuck with a function of a function?

I think I use the chain rule dy/dx = dy/du X du/dx

But I don't know what to do. Could someone explain it to me really simply please?
Say you have y = sin(4x).

A variation of the chain rule is to let u = 4x, v = sin(u).

dy/dx=dv/dx=dv/du×du/dx=cos(u)×4=4cos(u)=4cos(4x)dy/dx = dv/dx = dv/du \times du/dx = cos(u) \times 4 = 4cos(u) = 4cos(4x)

Does that make sense?
Reply 2
Avatar for raheem94
raheem94
13
Original post by jami74
I'm stuck with a function of a function?

I think I use the chain rule dy/dx = dy/du X du/dx

But I don't know what to do. Could someone explain it to me really simply please?


Its better that you give the question you are stuck on.
Reply 3
Avatar for jami74
jami74
OP
12
Original post by ViralRiver
Say you have y = sin(4x).

A variation of the chain rule is to let u = 4x, v = sin(u).

dy/dx=dv/dx=dv/du×du/dx=cos(u)×4=4cos(u)=4cos(4x)dy/dx = dv/dx = dv/du \times du/dx = cos(u) \times 4 = 4cos(u) = 4cos(4x)

Does that make sense?


I'm looking at it very hard. If I keep looking it might make sense and then I can apply it to my question.


Original post by raheem94
Its better that you give the question you are stuck on.


Thank-you. It is y=cos^2x

So is u = cos^2?
Reply 4
Avatar for raheem94
raheem94
13
Original post by jami74
I'm looking at it very hard. If I keep looking it might make sense and then I can apply it to my question.




Thank-you. It is y=cos^2x

So is u = cos^2?


y=cos2xu=cosxy=u2 y = cos^2x \\ u=cosx \\ y=u^2

Now find dydu  and  dudx \displaystyle \frac{dy}{du} \ \ and \ \ \frac{du}{dx} .

At the end, use the chain rule, dydx=dydu×dudx \displaystyle \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}
Reply 5
Avatar for jami74
jami74
OP
12
Original post by raheem94
y=cos2xu=cosxy=u2 y = cos^2x \\ u=cosx \\ y=u^2

Now find dydu  and  dudx \displaystyle \frac{dy}{du} \ \ and \ \ \frac{du}{dx} .

At the end, use the chain rule, dydx=dydu×dudx \displaystyle \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}


Thank-you.

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