Implicit differentiation

[Guest]

Is there anyone who can explain this in a dummied down way for me :shock: Thanks.

 

[daveyp225]

It is the same as regular differentiation except not in the form y = f(x).

For a function f(x,y), meaning some function with both x's and y's, Everytime you take the derivative of a term containing a y, you write y' (or dy/dx).

For example,

d/dx [ (x^2) + 2y = 3(y)^3 ]
-> 2x + 2y' = 9[(y)^2]y'

Notice how I took the derivative as normal, but being careful to use the chain rule. Y itself is a function and you must take the derivative of it as an "inside" function.

Here are some more examples:

d/dx [ y^3 = 7 ] -> (3y^2)(y') = 0
d/dx [ 2x + 3y = 5y ] -> 2 + 3y' = 5y'
d/dt [ 4p^2 + 3q = q ] -> 8p(dp/dt) + 3(dq/dt) = dq/dt

Try doing these:

d/dx [ 8y^2 + sin(y) = cos(yx) ]
d/dt [m^2 + n^3 = 3t ]

 

[Guest]

This is a problem that I was trying to do with implicit differentiation.

(2x-y)^4+2y^3=83.

I distributed the power to both the x's and y's then took the derivative getting:

8x^3-4y^3(dy/dx)+6y(dy/dx)=0

(dy/dx) (-4y^3+6y)=-8x^3.

(dy/dx)=-8x^3
----------
(-4y^3+6y)

I had to find y prime at (2,1) and then the tangent line.

I plugged in (2,1) and got -32/1

And for the tangent line : y-1=-32(x-2) or y=-32x+65

When I turned these in they all came back to me as wrong, can anyone explain to me what I did wrong?

 

[pka]

For the expression \(\displaystyle \L
\left( {2x - y} \right)^4 + 2y^3 = 83\) the implicit derivative is: \(\displaystyle \L
4(2x - y)^3 \left( {2 - y'} \right) + 6y^2 y' = 0\).

Now you solve y’.

 

[galactus]

Here's what I done:

\(\displaystyle (2x-y)^{4}+2y^{3}=83\)

Don't forget the chain rule.

\(\displaystyle 4(2x-y)^{3}(2-\frac{dy}{dx})+6y^{2}\frac{dy}{dx}=0\)

\(\displaystyle 8(2x-y)^{3}-4(2x-y)^{3}\frac{dy}{dx}+6y^{2}\frac{dy}{dx}=0\)

\(\displaystyle \frac{dy}{dx}=\frac{-8(2x-y)^{3}}{6y^{2}-4(2x-y)^{3}}\)

Subbing in x=2 and y=1, we get \(\displaystyle \frac{36}{17}\)

 

[Guest]

I'm really confused. Where did you get the 8 in the second step?

 

[pka]

Try to work through my posting.
Then you will see.

 

[Guest]

Well, I did, and I understood how you got that derivative. I just got confused on the last step where the 8 is placed out front. Did you just distribute? I tried using the product rule but when I did that I got a different answer.

 

[Guest]

ok, so in another problem, (2+y)^3+2y=x+69 I got this far:

3(2+y)^2(dy/dx)+2(dy/dx)=1

I don't know where to go after this, I tried finding it in that other problem but didn't understand.

 

[daveyp225]

3(2+y)^2(dy/dx)+2(dy/dx)=1

If your teacher wants you to solve for dy/dx, just factor it out:

(dy/dx)(3(2+y)^2+2)=1

Now, divide both sides by (3(2+y)^2+2):

dy/dx = (3*(2+y)^2+2)^-1

 

[galactus]

I'm really confused. Where did you get the 8 in the second step?

Hello:

I got the 8 from using the chain rule.

\(\displaystyle 4(2x-y)^{3}(2)=8(2x-y)^{3}\)