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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.

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  • $\begingroup$ Does "cbrt" mean "cube root?" $\endgroup$ – saulspatz Feb 28 '19 at 22:30
  • $\begingroup$ Yes it does, I didn't know if that or ^(1/3) would be more legible $\endgroup$ – jl8n Feb 28 '19 at 22:33
  • $\begingroup$ 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. $\endgroup$ – saulspatz Feb 28 '19 at 22:40
  • $\begingroup$ Thanks I appreciate it $\endgroup$ – jl8n Feb 28 '19 at 22:42
  • $\begingroup$ Why do you say you get an imaginary number? $(-1)^{2/3}=1.$ $\endgroup$ – saulspatz Feb 28 '19 at 22:42
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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.

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