lesson 6 homework: problem 19 (1 point)\nfind the equation of the tangent line to the curve $y =…

lesson 6 homework: problem 19 (1 point)\nfind the equation of the tangent line to the curve $y = -\\ln(x^{3}-26)$ at the point $(3,0)$.\n$y = $\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have 3 attempts remaining.\nemail instructor

lesson 6 homework: problem 19 (1 point)\nfind the equation of the tangent line to the curve $y = -\\ln(x^{3}-26)$ at the point $(3,0)$.\n$y = $\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have 3 attempts remaining.\nemail instructor

Answer

Explanation:

Step1: Find the derivative of the function

Use the chain - rule. If $y =-\ln(u)$ where $u=x^{3}-26$, then $\frac{dy}{du}=-\frac{1}{u}$ and $\frac{du}{dx}=3x^{2}$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=-\frac{3x^{2}}{x^{3}-26}$.

Step2: Evaluate the derivative at the given point

Substitute $x = 3$ into $\frac{dy}{dx}$. When $x = 3$,[Client Connection Error]