Posted by **Melissa** on Thursday, March 14, 2013 at 1:57pm.

Use implicit differentiation to find the slope of the tangent line to the curve y/( x + 4 y) = x^4 – 4

at the point (1,-3/13)

- Math -
**Steve**, Thursday, March 14, 2013 at 3:17pm
y/(x+4y) = x^4 – 4

(y'(x+4y) - y(1+4y'))/(x+4y)^2 = 4x^3

(x+4y-4y)y' = 4x^3 (x+4y)^2 +y

y' = 4x^2 (x+4y)^2 + y/x

y'(1) = 4(1)(1-12/13) + (-3/13)/1 = 1/13

- Math -
**Melissa**, Thursday, March 14, 2013 at 3:27pm
Steve,

That still isn't the correct answer :(

- Math -
**Steve**, Thursday, March 14, 2013 at 4:00pm
Hmmm. Have you any clues? what is the correct answer? see any mistakes in my agebra?

If you go to wolframalpha.com and type in

derivative y/(x+4y) = x^4 – 4

you will see the value of y', which is what I got above. So, if my answer is wrong, I must have erred in my evaluation at (1, -3/13).

- Math -
**Steve**, Thursday, March 14, 2013 at 4:03pm
Ah. I see I didn't take (1-12/13)^2

You can fix that.

- Math -
**Reiny**, Thursday, March 14, 2013 at 4:12pm
I first changed the equation to:

y = x^5 - 4x + 4x^4y - 16y

now differentiate ....

y' = 5x^4 - 4 + (4x^4)y' + 16x^3 y' - 16y'

y'( 1 - 4x^4 + 16) = 5x^4 - 4 + 16x^3

using the point (1, 3/13)

y' (1-4-16) = 5 - 4 - 48/13

-19y' = -35/13

y' = 35/247

- Math -
**Steve**, Thursday, March 14, 2013 at 4:39pm
oops: 16x^3 y' ?

- Math -
**Reiny**, Thursday, March 14, 2013 at 6:06pm
yup, that's the one

should have been 16x^3y

should have realized there were too many y' hanging around

Thanks

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