If x = int_0^y frac{1}{sqrt{1 + 9t^2}} dt and | vedclass

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

Question

(C) Given $x = \int_0^y \frac{1}{\sqrt{1 + 9t^2}} dt$. By the Fundamental Theorem of Calculus, differentiating both sides with respect to $y$, we get

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) Given $x = \int_0^y \frac{1}{\sqrt{1 + 9t^2}} dt$. By the Fundamental Theorem of Calculus, differentiating both sides with respect to $y$, we get $\frac{dx}{dy} = \frac{1}{\sqrt{1 + 9y^2}}$. Taking the reciprocal, we have $\frac{dy}{dx} = \sqrt{1 + 9y^2}$. Differentiating both sides with respect to $x$ using the chain rule, we get $\frac{d^2y}{dx^2} = \frac{d}{dx}(\sqrt{1 + 9y^2}) = \frac{1}{2\sqrt{1 + 9y^2}} \cdot (18y) \cdot \frac{dy}{dx}$. Substituting $\frac{dy}{dx} = \sqrt{1 + 9y^2}$, we get $\frac{d^2y}{dx^2} = \frac{9y}{\sqrt{1 + 9y^2}} \cdot \sqrt{1 + 9y^2} = 9y$. Comparing $\frac{d^2y}{dx^2} = 9y$ with $\frac{d^2y}{dx^2} = ay$, we find $a = 9$.

If $x = \int_0^y \frac{1}{\sqrt{1 + 9t^2}} dt$ and $\frac{d^2y}{dx^2} = ay$, then $a$ is equal to

Similar Questions

Let $p(x)$ be a polynomial such that $p(x) - p'(x) = x^n$,where $n$ is a positive integer. Then,$p(0)$ equals

$\frac{d^n}{dx^n}(\log x) =$

If $y^2 = a x^2 + 2 x + c$,then $y^3 \frac{d^2 y}{d x^2}$ is

If $y = (1 - x^2) \operatorname{Tanh}^{-1} x$, then $\frac{d^2 y}{d x^2} = $

If $f(x) = \begin{cases} x^3 \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0 \end{cases}$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module