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If x^(3)+4-5x^(2)=-5y+y^(2) then find (dy)/(dx) at the point (1,5). Answer: (dy)/(dx)|_((1,5))=

If x3+4−5x2=−5y+y2 x^{3}+4-5 x^{2}=-5 y+y^{2} then find dydx \frac{d y}{d x} at the point (1,5) (1,5) .\newlineAnswer: dydx∣(1,5)= \left.\frac{d y}{d x}\right|_{(1,5)}=

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Q. If x3+4−5x2=−5y+y2 x^{3}+4-5 x^{2}=-5 y+y^{2} then find dydx \frac{d y}{d x} at the point (1,5) (1,5) .\newlineAnswer: dydx∣(1,5)= \left.\frac{d y}{d x}\right|_{(1,5)}=
  1. Implicit Differentiation: To find the derivative of yy with respect to xx, dydx\frac{dy}{dx}, we need to implicitly differentiate both sides of the equation with respect to xx.\newlineGiven equation: x3+4−5x2=−5y+y2x^{3} + 4 - 5x^{2} = -5y + y^{2}\newlineDifferentiate both sides with respect to xx:\newlineddx[x3+4−5x2]=ddx[−5y+y2]\frac{d}{dx} [x^{3} + 4 - 5x^{2}] = \frac{d}{dx} [-5y + y^{2}]\newlineUsing the power rule and chain rule, we get:\newline3x2−10x=−5dydx+2ydydx3x^{2} - 10x = -5\frac{dy}{dx} + 2y\frac{dy}{dx}
  2. Isolating dydx\frac{dy}{dx}: Now we need to solve for dydx\frac{dy}{dx} by isolating it on one side of the equation.\newline3x2−10x=−5dydx+2ydydx3x^{2} - 10x = -5\frac{dy}{dx} + 2y\frac{dy}{dx}\newlineCombine like terms:\newlinedydx(2y−5)=3x2−10x\frac{dy}{dx}(2y - 5) = 3x^{2} - 10x\newlineNow, divide both sides by (2y−5)(2y - 5) to solve for dydx\frac{dy}{dx}:\newlinedydx=3x2−10x2y−5\frac{dy}{dx} = \frac{3x^{2} - 10x}{2y - 5}
  3. Evaluate at (1,5)(1,5): We need to evaluate dydx\frac{dy}{dx} at the point (1,5)(1,5).\newlineSubstitute x=1x = 1 and y=5y = 5 into the equation:\newlinedydx=3(1)2−10(1)2(5)−5\frac{dy}{dx} = \frac{3(1)^{2} - 10(1)}{2(5) - 5}\newlineSimplify the equation:\newlinedydx=3−1010−5\frac{dy}{dx} = \frac{3 - 10}{10 - 5}\newlinedydx=−75\frac{dy}{dx} = \frac{-7}{5}

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