The slope of the normal to the curve y=frac{x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The slope of the tangent to the curve $y=f(x)$ is given by $\frac{dy}{dx}$.Given $y = \frac{x}{x^2+1}$.Differentiating with respect to $x$ using t

Vedclass · Accepted Answer

$\frac{289}{15}$

Vedclass · Answer

(C) The slope of the tangent to the curve $y=f(x)$ is given by $\frac{dy}{dx}$.Given $y = \frac{x}{x^2+1}$.Differentiating with respect to $x$ using the quotient rule:$\frac{dy}{dx} = \frac{(x^2+1)(1) - x(2x)}{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2}$.At $x = -4$,the slope of the tangent is:$\left(\frac{dy}{dx}ight)_{x=-4} = \frac{1-(-4)^2}{((-4)^2+1)^2} = \frac{1-16}{(16+1)^2} = \frac{-15}{17^2} = \frac{-15}{289}$.The slope of the normal is the negative reciprocal of the slope of the tangent:$	ext{Slope of normal} = -\frac{1}{\left(\frac{dy}{dx}ight)} = -\frac{1}{\left(\frac{-15}{289}ight)} = \frac{289}{15}$.

The slope of the normal to the curve $y=\frac{x}{x^2+1}$ at $x=-4$ is

Similar Questions

The angle between the curves $y=\sin x$ and $y=\cos x$ is

Show that the equation of the normal at any point on the curve $x=3 \cos \theta-\cos ^{3} \theta, y=3 \sin \theta-\sin ^{3} \theta$ is $4(y \cos ^{3} \theta-x \sin ^{3} \theta)=3 \sin 4 \theta$.

Find the equation of the tangent line to the curve $y=x^{2}-2x+7$ which is perpendicular to the line $5y-15x=13$.

The tangent to a non-linear curve $y = f(x)$ at any point $P(x, y)$ intersects the $x$-axis and $y$-axis at $A$ and $B$ respectively. If the normal to the curve $y = f(x)$ at $P$ intersects the $y$-axis at $C$ such that $AC = BC$,and $f(2) = 3$,then the equation of the curve is:

The equation of the normal to the curve $y^4=a x^3$ at $(a, a)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module