If y = x - x^2,then the rate of change of y^2 | vedclass

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

Question

(C) Given $y = x - x^2$. We need to find the rate of change of $y^2$ with respect to $x^2$. Let $v = y^2 = (x - x^2)^2 = x^2 + x^4 - 2x^3$. Let $u = x

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Given $y = x - x^2$. We need to find the rate of change of $y^2$ with respect to $x^2$. Let $v = y^2 = (x - x^2)^2 = x^2 + x^4 - 2x^3$. Let $u = x^2$. We want to calculate $\frac{dv}{du} = \frac{dv/dx}{du/dx}$. First,differentiate $v$ with respect to $x$: $\frac{dv}{dx} = \frac{d}{dx}(x^2 + x^4 - 2x^3) = 2x + 4x^3 - 6x^2$. Next,differentiate $u$ with respect to $x$: $\frac{du}{dx} = \frac{d}{dx}(x^2) = 2x$. Now,$\frac{dv}{du} = \frac{2x + 4x^3 - 6x^2}{2x} = 1 + 2x^2 - 3x$. At $x = 2$,the rate of change is $1 + 2(2)^2 - 3(2) = 1 + 8 - 6 = 3$.

If $y = x - x^2$,then the rate of change of $y^2$ with respect to $x^2$ at $x = 2$ is

Similar Questions

The radius of a circle is increasing uniformly at the rate of $3 \, cm/s$. The rate of increase of its area when the radius is $10 \, cm$ will be:

$A$ stone thrown vertically upwards rises $s$ metres in $t$ seconds,where $s = 80t - 16t^2$. Then,the velocity after $2$ seconds is .......... $m/s$.

If a particle moves in a straight line according to the law $x = a \sin (\sqrt{\lambda} t + b)$, then the particle will come to rest at two points whose distance is [symbols have their usual meaning]

Water is being poured into a cylindrical vessel with a radius of $2 \text{ cm}$ at a rate of $8 \text{ cm}^3/\text{s}$. The rate at which the height of the water is increasing is:

If $r$ is the radius of a spherical balloon at time $t$ and the surface area of the balloon changes at a constant rate $K,$ then ....

Vedclass Test Series

Exam Paper Generator

Online Exam Module