## Calculus: Early Transcendentals 8th Edition

1a. $y'=\frac{9x}{y}$ 1b. $y'=\pm\dfrac{9x}{\sqrt{9x^2-1}}$ 1c. See solution below.
1a. $9x^2-y^2=1$. Taking the derivative of both sides we get $18x-2yy'=0$. Solving for $y'$ we get $y'=\frac{9x}{y}$. 1b. Solving for $y$ we get $y^2=9x^2-1$ so $y=\pm\sqrt{ 9x^2-1}$. Then $y'=\pm\frac{1}{2}(9x^2-1)^{-\frac{1}{2}}*18x$ using chain rule. Simplifying, we get $y'=\pm\dfrac{9x}{\sqrt{9x^2-1}}$. 1c. We can see that $y'=\pm\dfrac{9x}{\sqrt{9x^2-1}}=\dfrac{9x}{y}$ since $y$ from part b is $\pm\sqrt{ 9x^2-1}$ which checks out with our answer in part a.