We are given that (dy)/(dx)=sqrt(1-y^(2))". " Find an expression for (d^(2)y)/(dx^(2)) in terms of x and y. (d^(2)y)/(dx^(2))=

Given Derivative: We are given the first derivative of y with respect to x as (dy)/(dx) = sqrt(1 - y^2). To find the second derivative, we need to differentiate (dy)/(dx) with respect to x.Apply Chain Rule: Using the chain rule, we differentiate (dy)/(dx) with respect to x. Since y is a function of x, we treat y as a function y(x) and apply the chain rule accordingly. (d^(2)y)/(dx^(2)) = d/dx (dy/dx) = d/dx (sqrt(1 - y^2))Differentiate sqrt(1 - y^2): To differentiate sqrt(1 - y^2), we use the chain rule again. Let u = 1 - y^2, then the derivative of sqrt(u) with respect to u is (1/2)u^(-1/2). (d/dx) (sqrt(1 - y^2)) = (d/dx) (sqrt(u)) * (du/dy) * (dy/dx)Find du/dy: Now we find du/dy, which is the derivative of u = 1 - y^2 with respect to y. du/dy = d/dy (1 - y^2) = -2ySubstitute and Simplify: Substitute du/dy into the previous expression and simplify. (d/dx) (sqrt(1 - y^2)) = (1/2)(1 - y^2)^(-1/2) * (-2y) * (dy/dx)Substitute dy/dx: We already know that (dy/dx) = sqrt(1 - y^2), so we substitute this into the expression. (d^(2)y)/(dx^(2)) = (1/2)(1 - y^2)^(-1/2) * (-2y) * sqrt(1 - y^2)Simplify Expression: Simplify the expression by multiplying the terms together. (d^(2)y)/(dx^(2)) = -y(1 - y^2)^(-1/2) * (1 - y^2)^(1/2)Final Result: Since (1 - y^2)^(-1/2) * (1 - y^2)^(1/2) equals 1, the expression simplifies to: (d^(2)y)/(dx^(2)) = -y

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

We are given that

(dy)/(dx)=sqrt(1-y^(2))". "
Find an expression for
(d^(2)y)/(dx^(2)) in terms of
x and
y.

(d^(2)y)/(dx^(2))=

We are given that $\newline$ $\frac{d y}{d x}=\sqrt{1-y^{2}} \text {. }$ $\newline$ Find an expression for $\frac{d^{2} y}{d x^{2}}$ in terms of $x$ and $y$ . $\newline$ $\frac{d^{2} y}{d x^{2}}=$

View step-by-step help

Home

Math Problems

Algebra 1

Transformations of quadratic functions

Full solution

Q. We are given that

\newline

\frac{d y}{d x}=\sqrt{1-y^{2}} \text {. }

\newline

Find an expression for

\frac{d^{2} y}{d x^{2}}

in terms of

x

and

y

\newline

\frac{d^{2} y}{d x^{2}}=

Given Derivative: We are given the first derivative of $y$ with respect to $x$ as $\frac{dy}{dx} = \sqrt{1 - y^2}$ . To find the second derivative, we need to differentiate $\frac{dy}{dx}$ with respect to $x$ .
Apply Chain Rule: Using the chain rule, we differentiate $\frac{dy}{dx}$ with respect to $x$ . Since $y$ is a function of $x$ , we treat $y$ as a function $y(x)$ and apply the chain rule accordingly.\frac{d^{$2$}y}{dx^{$2$}} = \frac{d}{dx} \left(\frac{dy}{dx}\right) = \frac{d}{dx} \left(\sqrt{$1$ - y^{$2$}}\right)
Differentiate \(\sqrt{1 - y^2}: To differentiate $\sqrt{1 - y^2}$ , we use the chain rule again. Let $u = 1 - y^2$ , then the derivative of $\sqrt{u}$ with respect to $u$ is $(\frac{1}{2})u^{-\frac{1}{2}}$ . $\newline$ $\frac{d}{dx} (\sqrt{1 - y^2}) = \frac{d}{dx} (\sqrt{u}) \cdot \frac{du}{dy} \cdot \frac{dy}{dx}$
Find $\frac{du}{dy}$ : Now we find $\frac{du}{dy}$ , which is the derivative of $u = 1 - y^2$ with respect to $y$ . $\newline$ $\frac{du}{dy} = \frac{d}{dy} (1 - y^2) = -2y$
Substitute and Simplify: Substitute $\frac{du}{dy}$ into the previous expression and simplify. $\frac{d}{dx} (\sqrt{1 - y^2}) = \left(\frac{1}{2}\right)(1 - y^2)^{-\frac{1}{2}} \cdot (-2y) \cdot \left(\frac{dy}{dx}\right)$
Substitute $\frac{dy}{dx}$ : We already know that $\frac{dy}{dx} = \sqrt{1 - y^2}$ , so we substitute this into the expression. $\frac{d^2y}{dx^2} = \frac{1}{2}(1 - y^2)^{-\frac{1}{2}} \cdot (-2y) \cdot \sqrt{1 - y^2}$
Simplify Expression: Simplify the expression by multiplying the terms together. $\newline$ $\frac{d^{2}y}{dx^{2}} = -y(1 - y^{2})^{-\frac{1}{2}} \times (1 - y^{2})^{\frac{1}{2}}$
Final Result: Since $(1 - y^2)^{-\frac{1}{2}} \cdot (1 - y^2)^{\frac{1}{2}}$ equals $1$ , the expression simplifies to: $\frac{d^{2}y}{dx^{2}} = -y$