Solve Basic Calc Probs: dy/dx for y=z7, z=sin(x) & y=√z , z=x4+1

[ar202]

Homework Statement

for each of the following find dy/dx, expressing the answer as a function of x.

(A) y=z⁷, z=sin(x)

(B) y=√z , z=x⁴+1

Homework Equations

The Attempt at a Solution

I'm having some trouble working out what the question is actually asking me to do... If i go at it normally i get for eg 7z⁶cos(x) ... but this isn't the answer.

 

[Doc Al]

If i go at it normally i get for eg 7z⁶cos(x) ... but this isn't the answer.

Express the answer as a function of x. (Get rid of that z!)

 

[ar202]

so I'm literally just replacing the z?

sin(x)⁷sin(x)

becomes

7sin(x)⁶cos(x)

 

[lanedance]

the 2nd line looks ok, but why do you start with sin(x)⁷sin(x)?

also to help people help you, try & be clear in what you want to communicate and try and give all the info eg.

so substituting in for z(x) = sin(x) gives
y(x) = sin(x)⁷
then differentiating w.r.t. x
y'(x) = 7sin(x)⁶cos(x)

 

[ar202]

the 2nd line look ok, but why do you start with sin(x)⁷sin(x)?

i was replacing the z... prob mixed up the steps

Are these questions based on the chain rule?

 

[Doc Al]

so I'm literally just replacing the z?

Yes. Your answer of 7z⁶cos(x) becomes 7sin⁶(x)cos(x).

 

[Doc Al]

Are these questions based on the chain rule?

Yes.

 

[lanedance]

Are these questions based on the chain rule?

yes & and that's exactly what you've used

 

[ar202]

ok cool, because i was doing some work on products and quotients and then got hit with these so was a little thrown.

thanks for your help!

ps Any help with (B) :p

 

[lanedance]

in essence same as a) using chain rule

 

[lanedance]

in both cases you are given y(z) and z(x), then substitute to get y(z(x))

the derivative of y w.r.t. the chain rule is
$$\frac{dy(z(x))}{dx} = \frac{dy(z)}{dz} \frac{dz(x)}{dx}$$

noting as you did early on you need to substitute in for z = z(x) in the final answer if any remain

 

[ar202]

ok I am kinda struggling with the latex code... what a ballache! apologies if this is confusing

z=1/2√z & z=x⁴+1 = 4x³

then i get

4x³/2√x⁴+1

how does that look...?

 

[Mark44]

ok I am kinda struggling with the latex code... what a ballache! apologies if this is confusing

z=1/2√z & z=x⁴+1 = 4x³

It should be y = (1/2)sqrt(z) + 1. You don't want z to be defined as the square root of itself. Then, if z = z=x⁴+1, dz/dx = 4x³, not z.

then i get

4x³/2√x⁴+1

how does that look...?

Other than it's incorrect? It should be written as an equation that identifies what you have found.

You might find it easier to work with exponents rather than radicals. y = (1/2)z^1/2. dy/dx = dy/dz * dz/dx

 

[ar202]

It should be y = (1/2)sqrt(z) + 1. You don't want z to be defined as the square root of itself. Then, if z = z=x⁴+1, dz/dx = 4x³, not z.

Other than it's incorrect? It should be written as an equation that identifies what you have found.

You might find it easier to work with exponents rather than radicals. y = (1/2)z^1/2. dy/dx = dy/dz * dz/dx

ok. thanks for the help dude