how do i find dy/dx? where y = cos (xy^2) could i say dy/dx = -sin (xy^2) times the derivative of the inside stuff? with respect to x. i suppose i would use the product rule? right or is there an easier way to do this problem?
Hello, slapmaxwell1!
How do i find $\displaystyle \frac{dy}{dx}\,\text{ where }\,y \:=\: \cos(xy^2)$
Could i say: .$\displaystyle \frac{dy}{dx} \:=\: -\sin(xy^2)$ .times the derivative of the inside stuff with respect to x? . Yes!
i suppose i would use the product rule, right? . Right!
or is there an easier way to do this problem? . I don't think so.
You'd have: .$\displaystyle \frac{dy}{dx} \;=\;-\sin(xy^2)\cdot\bigg[x\cdot2y\frac{dy}{dx} + y^2\bigg] $
Then solve for $\displaystyle \frac{dy}{dx}$ . . . carefully.