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
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