Practice problem
Find dydx\frac{dy}{dx}dxdy for tan(xy)=x2+3y2\tan\bigg(\frac xy\bigg)=x^2+3y^2tan(yx)=x2+3y2
dydx=sec2(xy)−xy2x2sec2(xy)+2y2\frac{dy}{dx}=\frac{\sec^2(\frac xy)-xy^2}{x^2\sec^2(\frac xy)+2y^2}dxdy=x2sec2(yx)+2y2sec2(yx)−xy2
dydx=y(sec2(xy)−2xy)xsec2(xy)+6y3\frac{dy}{dx}=\frac{y(\sec^2(\frac xy)-2xy)}{x\sec^2(\frac xy)+6y^3}dxdy=xsec2(yx)+6y3y(sec2(yx)−2xy)
dydx=sec2(xy)+4x3yxysec2(xy)−4y\frac{dy}{dx}=\frac{\sec^2(\frac xy)+4x^3y}{xy\sec^2(\frac xy)-4y}dxdy=xysec2(yx)−4ysec2(yx)+4x3y
dydx=xy(2sec2(xy)−2y)sec2(xy)+7xy\frac{dy}{dx}=\frac{xy(2\sec^2(\frac xy)-2y)}{\sec^2(\frac xy)+7x\sqrt y}dxdy=sec2(yx)+7xyxy(2sec2(yx)−2y)
Type a concept, like “slope” or “photosynthesis”.
