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Question Number 163367 by poeter last updated on 06/Jan/22

Answered by som(math1967) last updated on 06/Jan/22

$$1.equnofcircle{x}^2+y^2=10\left(given\right)$$$$\frac{dy}{dx}=\frac{-x}{y}$$$${\left[\frac{dy}{dx}\right]}_{1,3}=\frac{-1}{3}$$$$\therefore slopeoftanjent=\frac{-1}{3}$$$$equationoftanjent$$$$(y-3)=\frac{-1}{3}(x-1)$$$$x+3y=10$$

Answered by MikeH last updated on 06/Jan/22

$$f(x,y)=4x^3{y}^2+6y^3$$$$\frac{\partial f}{\partial x}=12x^2{y}^2+0=12x^2{y}^2$$$$\frac{\partial f}{\partial y}=8x^{3}y+18y^2$$

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