Which of the following points lies on the tan | vedclass

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(B) Given the curve equation: $x^{4} e^{y}+2 \sqrt{y+1}=3$. Differentiating both sides with respect to $x$: $x^{4} e^{y} y^{\prime} + 4x^{3} e^{y} + \

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$(-2,6)$

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(B) Given the curve equation: $x^{4} e^{y}+2 \sqrt{y+1}=3$. Differentiating both sides with respect to $x$: $x^{4} e^{y} y^{\prime} + 4x^{3} e^{y} + \frac{2 y^{\prime}}{2 \sqrt{y+1}} = 0$. At the point $P(1,0)$,substitute $x=1$ and $y=0$: $(1)^{4} e^{0} y^{\prime} + 4(1)^{3} e^{0} + \frac{y^{\prime}}{\sqrt{0+1}} = 0$. $y^{\prime} + 4 + y^{\prime} = 0$. $2y^{\prime} = -4$,which gives $y^{\prime} = -2$. The equation of the tangent at $P(1,0)$ with slope $m = -2$ is: $y - 0 = -2(x - 1)$. $y = -2x + 2$,or $2x + y = 2$. Checking the options: For $(-2, 6)$,$2(-2) + 6 = -4 + 6 = 2$. Thus,the point $(-2, 6)$ lies on the tangent line.

Which of the following points lies on the tangent to the curve $x^{4} e^{y}+2 \sqrt{y+1}=3$ at the point $(1,0)$?

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