implicit differentiation help?

How do i differentiate
dy/dx(1+x)
with respect to x?

what i posted is the first derivative and im trying to find the second derivative but having difficulties on how to differentiate it again

Reply 1

ErniePicks

OP

2

Original post by zetamcfc

Product rule.

Spoiler

So U=d/dx and v=dy/dx * (1+x) ?
I'm not sure what the derivatives of d/dx and dy/dx are though...

Reply 2

Zacken

22

Original post by ErniePicks

How do i differentiate
dy/dx(1+x)
with respect to x?

what i posted is the first derivative and im trying to find the second derivative but having difficulties on how to differentiate it again

Use the product rule.

\frac{d}{dx}\left(\frac{dy}{dx} \right)(1+x) + \frac{d}{dx}(1+x) \frac{dy}{dx}

.

Remember that d/dx(dy/dx) = d^2y/dx^2. i.e: the derivative of the derivative is the second derivative.

(edited 7 years ago)

Reply 3

ErniePicks

OP

2

whoops nvm i did it u=dy/dx and v=(1+x)

so the derivative wrt x of dy/dx is d^2y/dx^2 but what is the derivative wrt to x for d/dx?

Reply 4

Zacken

22

Original post by ErniePicks

whoops nvm i did it u=dy/dx and v=(1+x)

so the derivative wrt x of dy/dx is d^2y/dx^2 but what is the derivative wrt to x for d/dx?

Well you tell me, surely you're used to differentiating

x^1

? It's just

1 \times x^{1-1}

.

Reply 5

zetamcfc

19

Original post by ErniePicks

So U=d/dx and v=dy/dx * (1+x) ?
I'm not sure what the derivatives of d/dx and dy/dx are though...

y is a function of x

\displaystyle K(x) =\frac{dy}{dx} (1+x)

\displaystyle \frac{d(K(x))}{dx} = \frac{d(\frac{dy}{dx} (1+x))}{dx}

\frac{d(K(x))}{dx} = \frac{d( \frac{dy}{dx} )}{dx} + \frac{d(x \frac{dy}{dx} )}{dx}

\frac{d(K(x))}{dx} = \frac{d^2 y}{dx^2} +( \frac{dy}{dx} + x\frac{d^2 y}{dx} )

\frac{d(K(x))}{dx} = (1+x) \frac{d^2 y}{dx} + \frac{dy}{dx}

Reply 6

ErniePicks

OP

2

cheers lads, got another bit that is confusing me. I have t differentiate wrt x

x^3+3y^2=x(y+x^2)

Now what i did was expand out the bracket with the x and then differentiate wrt x all of it and got the right answer. My teacher used a different method and got this though which i'm not sure how:

the RHS went to: x(dy/dx+2x)+y+x^2

Now i understand that the values inside the x bracket is what the original alues would be if they were differentiated wrt x but unsure about the y and x^2 comes from. Is my method of expanding out the brackets with the x and then differentiating okay to use?

Reply 7

Zacken

22

Original post by ErniePicks

Now i understand that the values inside the x bracket is what the original alues would be if they were differentiated wrt x but unsure about the y and x^2 comes from. Is my method of expanding out the brackets with the x and then differentiating okay to use?

Yes, your method is fine. The teacher has simply used the product rule with u = x and v = y + x^2

Reply 8

ErniePicks

OP

2

Original post by Zacken

Yes, your method is fine. The teacher has simply used the product rule with u = x and v = y + x^2

lol yeah i just realised this a few seconds ago

Reply 9

rosemondtan

10

Original post by ErniePicks

cheers lads, got another bit that is confusing me. I have t differentiate wrt x

x^3+3y^2=x(y+x^2)

Now what i did was expand out the bracket with the x and then differentiate wrt x all of it and got the right answer. My teacher used a different method and got this though which i'm not sure how:

the RHS went to: x(dy/dx+2x)+y+x^2

Now i understand that the values inside the x bracket is what the original alues would be if they were differentiated wrt x but unsure about the y and x^2 comes from. Is my method of expanding out the brackets with the x and then differentiating okay to use?

I'm guessing you could, either method should give you the same answer. I personally think product rule is faster though and in some cases you can't expand the brackets because it's too complicated and it makes it easier to make mistakes. Anyway here's how I did it

Oops I see you've realised why haha

implicit differentiation help?

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