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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Solution

$(y - x^{5})^{2} = x (1 + x^{2})^{2}$
$2 (y - x^{5}) (\frac{d y}{d x} - 5 x^{4}) = (1 + x^{2})^{2} + 2 x (1 + x^{2}) .2 x$
At point $(1, 3)$
$2 (3 - 1) (\frac{d y}{d x} - 5) = (1 + 1)^{2} + 2 (1 + 1) .2 \Rightarrow \frac{d y}{d x} = 8$

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Similar questions

Q. The slope of the tangent to the curve

(y - x^{5})^{2} = x (1 + x^{2})^{2}

at the point

(1, 3)

is ___

Q. The slope of the tangent to the curve

(y - x^{5})^{2} = x (1 + x^{2})^{2}

at the point

(1, 3)

is ..................

Q. The slope of the tangent to the curve

(y - x^{5})^{2} = x (1 + x^{2})^{2}

at the point

(1, 3)

is ___

Q. If

| f (x_{1}) - f (x_{2}) | < (x_{1} - x_{2})^{2}

for all

x_{1}

x_{2}

\in

R. Find the equation of tangent to the curve y = f(x) at the point (1, 2).

P and Q are 2 external points from which tangents are drawn to circle centered at origin and radius 'r' $P \equiv (x_{1}, y_{1}), Q \equiv (x_{2}, y_{2}) .$ What is the condition for the lengths of tangents to be the same?

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The slope of the tangent to the curve(y−x5)2=x(1+x2)2 at the point (1,3) is ___

(y−x5)2=x(1+x2)2 2(y−x5)(dydx−5x4)=(1+x2)2+2x(1+x2).2x At point (1,3) 2(3−1)(dydx−5)=(1+1)2+2(1+1).2⇒dydx=8

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

$(y - x^{5})^{2} = x (1 + x^{2})^{2}$
$2 (y - x^{5}) (\frac{d y}{d x} - 5 x^{4}) = (1 + x^{2})^{2} + 2 x (1 + x^{2}) .2 x$
At point $(1, 3)$
$2 (3 - 1) (\frac{d y}{d x} - 5) = (1 + 1)^{2} + 2 (1 + 1) .2 \Rightarrow \frac{d y}{d x} = 8$