# Implicit differentiation / find the equation of the tangent line using the derivative

So the first step in this problem is to find y' implicitly.

$$\sqrt[3]{y}=x^4-3$$

I managed to get that into this form

$$y'=12x^3y^{2/3}$$

I then need to find the equation of the tangent line at point $$(-1, -8),$$ but I cannot plug $$-1$$ into the derivative using point slope because y is on both sides. Simply plugging in $$x$$ and $$y$$ gives me an imaginary number which I don't believe is right. I'm definitely stuck and could use any advice, thanks in advance.

• Does "cbrt" mean "cube root?" – saulspatz Feb 28 '19 at 22:30
• Yes it does, I didn't know if that or ^(1/3) would be more legible – jl8n Feb 28 '19 at 22:33
• You should use MathJax to format questions on this site. You'll get a lot more positive response if your questions are easy to read. Start by clicking the edit button and looking at how I modified your post. – saulspatz Feb 28 '19 at 22:40
• Thanks I appreciate it – jl8n Feb 28 '19 at 22:42
• Why do you say you get an imaginary number? $(-1)^{2/3}=1.$ – saulspatz Feb 28 '19 at 22:42

Just write $$y'=12x^3\sqrt[3]{y^2}.$$
Now, the slope it's $$y'(-1,-8)=12(-1)^3\sqrt[3]{(-8)^2}=-48$$ and the rest is smooth.