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If 0=4+5x^(3)-y^(2) then find (dy)/(dx) at the point (1,-3). Answer: (dy)/(dx)|_((1,-3))=

If 0=4+5x3−y2 0=4+5 x^{3}-y^{2} then find dydx \frac{d y}{d x} at the point (1,−3) (1,-3) .\newlineAnswer: dydx∣(1,−3)= \left.\frac{d y}{d x}\right|_{(1,-3)}=

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Evaluate a nonlinear functionright chevron icon

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Q. If 0=4+5x3−y2 0=4+5 x^{3}-y^{2} then find dydx \frac{d y}{d x} at the point (1,−3) (1,-3) .\newlineAnswer: dydx∣(1,−3)= \left.\frac{d y}{d x}\right|_{(1,-3)}=
  1. Implicit Differentiation: First, we need to implicitly differentiate the given equation with respect to xx to find dydx\frac{dy}{dx}.\newlineGiven equation: 0=4+5x3−y20 = 4 + 5x^3 - y^2\newlineDifferentiate both sides with respect to xx:\newlineddx(0)=ddx(4)+ddx(5x3)−ddx(y2)\frac{d}{dx}(0) = \frac{d}{dx}(4) + \frac{d}{dx}(5x^3) - \frac{d}{dx}(y^2)\newlineSince the derivative of a constant is 00, we have:\newline0=0+15x2−2ydydx0 = 0 + 15x^2 - 2y\frac{dy}{dx}
  2. Solving for dydx\frac{dy}{dx}: Now, we solve for dydx\frac{dy}{dx}:
    0=15x2−2ydydx0 = 15x^2 - 2y\frac{dy}{dx}
    To isolate dydx\frac{dy}{dx}, we add 2ydydx2y\frac{dy}{dx} to both sides and then divide by 2y2y:
    2ydydx=15x22y\frac{dy}{dx} = 15x^2
    dydx=15x22y\frac{dy}{dx} = \frac{15x^2}{2y}
  3. Substitute Given Point: Next, we substitute the given point (1,−3)(1, -3) into the equation to find the value of dydx\frac{dy}{dx} at that point:\newline(dydx)∣(1,−3)=15(1)22(−3)\left(\frac{dy}{dx}\right)|_{(1,-3)} = \frac{15(1)^2}{2(-3)}\newline(dydx)∣(1,−3)=15−6\left(\frac{dy}{dx}\right)|_{(1,-3)} = \frac{15}{-6}\newline(dydx)∣(1,−3)=−52\left(\frac{dy}{dx}\right)|_{(1,-3)} = -\frac{5}{2}

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