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December 20, 2014

December 20, 2014

Posted by **Tyler** on Tuesday, June 25, 2013 at 5:25pm.

y^2=x^3(4−x)

at the point (2,16−−).

a. Find dy/dx at x=2.

dy/dx=

b. Write the equation of the tangent line to the curve.

- Calculus -
**Steve**, Tuesday, June 25, 2013 at 5:37pmy^2 = 4x^3 - x^4

2yy' = 12x^2 - 4x^3

y' = (6x^2 - 2x^3)/y

y(2) = 4

y'(2) = (6*4-2*8)/4 = 2

So, now you have a point and a slope. The tangent line at (2,4) is

y-4 = 2(x-2)

- Calculus -
**Reiny**, Tuesday, June 25, 2013 at 5:41pm2y dy/dx = x^3 (-1) + 3x^2 (4-x)

dy/dx = (-x^3 + 12x^2 - 3x^3)/(2y)

= (6x^2 - 2x^3)/y

I can't make out your point (2, 16−−)

but it should be easy for you

just plug in x=2 and y = whatever into the dy/dx

and that becomes your slope

then use the grade 9 way you learned to find the equation of a line with a given slope and a given point.

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