Differentiating twice with respect to x

[sooty1892]

Homework Statement

Differentiate twice: z = sinx

Homework Equations

Product rule
Chain rule

The Attempt at a Solution

dy/dx = dy/dz * dz/dx

dy/dx = dy/dz * cosx

Using the product rule:

d^2y/dx^2 = d^2y/dz^2 * cosx - dy/dz * sinx

According to the answer in the book the answer is: d^2y/dx^2 = d^2y/dz^2 * cos^2x - dy/dz * sinx but I don't see how.

Thanks for any help.

 

[Bacle2]

Why are you considering the product and chain? Product is used when you find the derivative of a product f.g , and chain is used when you have an expression f(g(x)),
and I don't see how you have either of these.

 

[sooty1892]

Ok. You're right that I don't use the chain rule but the product rule is used when differentiating dy/dx = dy/dz * cosx since there's 2 parts which both have variables.

 

[Ray Vickson]

Homework Statement

Differentiate twice: z = sinx

Homework Equations

Product rule
Chain rule

The Attempt at a Solution

dy/dx = dy/dz * dz/dx

dy/dx = dy/dz * cosx

Using the product rule:

d^2y/dx^2 = d^2y/dz^2 * cosx - dy/dz * sinx

According to the answer in the book the answer is: d^2y/dx^2 = d^2y/dz^2 * cos^2x - dy/dz * sinx but I don't see how.

Thanks for any help.

You need to be more careful when posing questions. The question you wrote has no y in it anywhere. Did you mean "find d^y/dx^2, when y = f(z) and z = sin(x)"?

RGV

 

[sooty1892]

I think I do. This is part of a much larger second order differential equations question where z = sin(x) is a substitution.