If y = at^2 + 2bt + c and t = ax^2 + 2bx + c, | vedclass

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

Question

(D) Given $y = at^2 + 2bt + c$ and $t = ax^2 + 2bx + c$. First,find $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{d

Vedclass · Accepted Answer

$24 a^2 (ax + b)$

Vedclass · Answer

(D) Given $y = at^2 + 2bt + c$ and $t = ax^2 + 2bx + c$. First,find $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$. $\frac{dy}{dt} = 2at + 2b = 2(at + b)$. $\frac{dt}{dx} = 2ax + 2b = 2(ax + b)$. So,$\frac{dy}{dx} = 2(at + b) \cdot 2(ax + b) = 4(at + b)(ax + b)$. Now,differentiate with respect to $x$: $\frac{d^2y}{dx^2} = 4 \left[ \frac{d}{dx}(at + b) \cdot (ax + b) + (at + b) \cdot \frac{d}{dx}(ax + b) \right]$. Since $\frac{dt}{dx} = 2(ax + b)$,we have $\frac{d}{dx}(at + b) = a \cdot \frac{dt}{dx} = 2a(ax + b)$. $\frac{d^2y}{dx^2} = 4 \left[ 2a(ax + b)^2 + (at + b) \cdot a \right] = 8a(ax + b)^2 + 4a(at + b)$. Finally,differentiate again to find $\frac{d^3y}{dx^3}$: $\frac{d^3y}{dx^3} = 8a \cdot 2(ax + b) \cdot a + 4a \cdot \frac{d}{dx}(at + b) = 16a^2(ax + b) + 4a \cdot 2a(ax + b) = 16a^2(ax + b) + 8a^2(ax + b) = 24a^2(ax + b)$.

If $y = at^2 + 2bt + c$ and $t = ax^2 + 2bx + c$,then $\frac{d^3y}{dx^3}$ equals

Similar Questions

Find the second order derivative of the function $y = e^{x} \sin 5x$.

If $y=a \sin x+(5+2 x) \cos x$,then $y^{\prime \prime}+y=$

If $y=5 \cos x-3 \sin x$,then $\frac{d^2 y}{d x^2}=$ . . . . . . .

If $x = \log p$ and $y = \frac{1}{p}$,then

If $y=2 \cos (2 \log x)+3 \sin (2 \log x)$, then $x^2 y^{\prime \prime}+x y^{\prime}+2 y=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module