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

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

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Q. If −5y+3y2+4+5x3=x2 -5 y+3 y^{2}+4+5 x^{3}=x^{2} then find dydx \frac{d y}{d x} at the point (−1,2) (-1,2) .\newlineAnswer: dydx∣(−1,2)= \left.\frac{d y}{d x}\right|_{(-1,2)}=
  1. Implicit Differentiation: First, we need to implicitly differentiate both sides of the equation with respect to xx. The equation is −5y+3y2+4+5x3=x2-5y + 3y^2 + 4 + 5x^3 = x^2.
  2. Differentiation with Chain Rule: Differentiate each term with respect to xx. For terms involving yy, we use the chain rule, treating yy as a function of xx (y=y(x)y = y(x)).\newlineddx(−5y)=−5dydx\frac{d}{dx}(-5y) = -5\frac{dy}{dx}, since yy is a function of xx.\newlineddx(3y2)=3⋅2y⋅dydx=6ydydx\frac{d}{dx}(3y^2) = 3 \cdot 2y \cdot \frac{dy}{dx} = 6y\frac{dy}{dx}, again using the chain rule.\newlineddx(4)=0\frac{d}{dx}(4) = 0, since the derivative of a constant is zero.\newlineyy00, using the power rule.\newlineyy11, using the power rule.
  3. Write Differentiated Equation: Now, we write down the differentiated equation:\newline−5dydx+6ydydx+0+15x2=2x-5\frac{dy}{dx} + 6y\frac{dy}{dx} + 0 + 15x^2 = 2x.
  4. Simplify Combined Terms: We can simplify the equation by combining like terms: (6y−5)dydx+15x2=2x. (6y - 5)\frac{dy}{dx} + 15x^2 = 2x.
  5. Solve for dydx\frac{dy}{dx}: Now, we solve for dydx\frac{dy}{dx}:dydx=2x−15x26y−5\frac{dy}{dx} = \frac{2x - 15x^2}{6y - 5}.
  6. Substitute Values: We substitute x=−1x = -1 and y=2y = 2 into the equation to find dydx\frac{dy}{dx} at the point (−1,2)(-1,2):(dydx)∣(−1,2)=(2(−1)−15(−1)2)(6(2)−5)\left(\frac{dy}{dx}\right)|_{(-1,2)} = \frac{(2(-1) - 15(-1)^2)}{(6(2) - 5)}.
  7. Calculate Result: Calculate the values:\newline(dydx)∣(−1,2)=(−2−15)(12−5).(\frac{dy}{dx})|_{(-1,2)} = \frac{(-2 - 15)}{(12 - 5)}.
  8. Final Simplification: Simplify the expression: \((\frac{dy}{dx})|_{(\(-1\),\(2\))} = \frac{\(-17\)}{\(7\)}\.

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