The slope of the tangent to the curve (y-x^5) | vedclass

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Question

(D) Given the curve equation: $(y-x^5)^2=x(1+x^2)^2$ Differentiating both sides with respect to $x$: $2(y-x^5)(\frac{dy}{dx}-5x^4) = (1+x^2)^2 + x \cd

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) Given the curve equation: $(y-x^5)^2=x(1+x^2)^2$ Differentiating both sides with respect to $x$: $2(y-x^5)(\frac{dy}{dx}-5x^4) = (1+x^2)^2 + x \cdot 2(1+x^2)(2x)$ $2(y-x^5)(\frac{dy}{dx}-5x^4) = (1+x^2)^2 + 4x^2(1+x^2)$ Substitute the point $(1,3)$ into the equation: $2(3-1^5)(\frac{dy}{dx}-5(1)^4) = (1+1^2)^2 + 4(1)^2(1+1^2)$ $2(3-1)(\frac{dy}{dx}-5) = (2)^2 + 4(2)$ $2(2)(\frac{dy}{dx}-5) = 4 + 8$ $4(\frac{dy}{dx}-5) = 12$ $\frac{dy}{dx}-5 = 3$ $\frac{dy}{dx} = 8$ Thus,the slope of the tangent at $(1,3)$ is $8$.

The slope of the tangent to the curve $(y-x^5)^2=x(1+x^2)^2$ at the point $(1,3)$ is

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