## University Calculus: Early Transcendentals (3rd Edition)

a) $\dfrac{1}{x^2}=y^2$ b) $\dfrac{1}{(x+3)^2}=y^2$ c) $\dfrac{1}{(x+C)^2}=y^2$
a) $y^2=\dfrac{1}{x^2}$ Thus, $y'=-[(-1) x^{-2}]=\dfrac{1}{x^2}=y^2$ b) $y^2=\dfrac{1}{(x+3)^2}$ Thus, $y'=-[(-1) (x+3)^{-2}]=\dfrac{1}{(x+3)^2}=y^2$ c) $y^2=\dfrac{1}{(x+C)^2}$ Thus, $y'=-[(-1) (x+C)^{-2}]=\dfrac{1}{(x+C)^2}=y^2$