Slope of Tangent Line at (0,-10) for y^3+1004=(e^x+1)^2

[muddyjch]

Homework Statement

A curve is given by the equation: y^3+1004=(e^x+1)^2
Find the slope of the tangent line at the point (0,-10).

Homework Equations

The Attempt at a Solution

I took the derivative of ((e^x+1)^2-1004)^(1/3) and that is (2e^x(1+e^x))/(3((1+e^x)^2-1004)^(2/3)) but plugging in 0 for x does not give me the right answer.

 

[Mentallic]

I've done it both by implicit differentiation and by your method and the answers both have given me 1/75. You probably just plugged in wrong!

 

[muddyjch]

Thanks that is the right answer, just didn't get into the calculator right.

 

[Mentallic]

There's no need for a calculator. Since you're dealing with exponentials and x=0, it is clear that the answer will be simple to solve by hand.

$$\frac{dy}{dx}=\frac{2e^x(e^x+1)}{3\left((e^x+1)^2-1004\right)^{2/3}}$$

x=0, $$\frac{dy}{dx}=\frac{2e^0(e^0+1)}{3\left((e^0+1)^2-1004\right)^{2/3}}$$

e^0=1 so this simplifies to $$\frac{dy}{dx}=\frac{4}{3(4-1004)^{2/3}}$$

Now $(-1000)^{1/3}=-10$ and $(-10)^2=100$ so $(4-1004)^{2/3}=100$.
That gets you the answer 1/75 as required, and no need for throwing a messy long expression into the calculator which, as you've seen, can easily lead to errors.