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Calculus/calculus-differentiation rules

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Expert: - 10/7/2009


Question
Suppose that
f (x) =2x^2/(5−4x)^4 .
Find an equation for the tangent line to the graph of f at x = 1.
Tangent line: y =

Any help would be great!

Answer
f'(x) = (lo d hi - hi d lo)/lo² = (((5-4x)^4)4x - 2x^2(4(5-4x)^3)(-4))/(5-4x)^8.
We can factor (5-4x)^3 out of the numerator and denominator.  This leaves
((5-4x)4x - 2x^2(4)(-4))/(5-4x)^5.

Multiplying out what is left in the numerator gives 20x - 16x² +32x².
-16+32=16, so in the numerator we get 20x + 16x².

Factor out an x and put the denominator back in to get x(20+16x)/(5-4x)^5.

Put x=1 into this equation and get 1(20+16)/(5-4)^5 = 36/1 = 36.
That says that the slope is 36.
The point is found by putting 1 into f(x) as x, giving 2/1 = 2.

The equation for the line is the y - 2 = 36(x-1).

Multiply out and add 2 to both sides, and you've got it.

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