Solved: Leibniz’s Rule In applications, we sometimes

Chapter 5, Problem 31AAE

(choose chapter or problem)

Get Unlimited Answers
QUESTION:

Leibniz’s Rule In applications, we sometimes encounter functions defined by integrals that have variable upper limits of integration and variable lower limits of integration at the same time. The first integral can be evaluated directly, but the second cannot. We may find the derivative of either integral, however, by a formula called Leibniz’s Rule.

Leibniz’s Rule

If f is continuous on \([a, b]\) and if \(u(x)\) and \(v(x)\) are differentiable functions of \(x\) whose values lie in \([a, b]\), then

\(\frac{d}{dx}\int_{u(x)}^{v(x)}f(t)dt=f(v(x))\frac{dy}{dx}-f(u(x))\frac{d u}{dx}\)

                                                 

To prove the rule, let \(F\) be an antiderivative of fon \([a, b]\). Then

\(\int_{\nu(x)}^{v(x)}f(t)d t=f(v(x))-F(u(x))\)              

                                       

Differentiating both sides of this equation with respect to \(x\) gives the equation we want:

\(\frac{d}{d x} \int_{u(x)}^{v(x)} f(t) d t=f(v(x)) \frac{d v}{d x}-f(u(x)) \frac{d u}{d x}\)

\(=F^{\prime}(v(x)) \frac{d v}{d x}-F^{\prime}(u(x)) \frac{d u}{d x}\)

\(=f(v(x)) \frac{d y}{\partial x}-f(u(x)) \frac{d u}{d x}\)        

                           

Use Leibniz's Rule to find the derivatives of the functions in Exercises 31-38.

        \(f(x)=\int_{1 / x}^{x} \frac{1}{t} d t\)

Equation Transcription:

Text Transcription:

f

[a,b]

u(x)

v(x)

x

[a,b]

d/dx integral over u(x) ^v(x) f(t)dt=f(v(x))dv/dx-f(u(x))du/dx

F

f

[a,b]

Integral over u(x) ^v(x) f(t)dt=f(v(x))-F(u(x))

x

d/dx integral over u(x) ^v(x) f(t)dt=d/dx[F(v(x))-F(u(x))]

=F'(v(x)dv/dx-F'(u(x))dudx

=f(v(x))dv/dx-f(u(x))dudx

f(x)= integral over 1/x ^x 1/t dt

Questions & Answers

QUESTION:

Leibniz’s Rule In applications, we sometimes encounter functions defined by integrals that have variable upper limits of integration and variable lower limits of integration at the same time. The first integral can be evaluated directly, but the second cannot. We may find the derivative of either integral, however, by a formula called Leibniz’s Rule.

Leibniz’s Rule

If f is continuous on \([a, b]\) and if \(u(x)\) and \(v(x)\) are differentiable functions of \(x\) whose values lie in \([a, b]\), then

\(\frac{d}{dx}\int_{u(x)}^{v(x)}f(t)dt=f(v(x))\frac{dy}{dx}-f(u(x))\frac{d u}{dx}\)

                                                 

To prove the rule, let \(F\) be an antiderivative of fon \([a, b]\). Then

\(\int_{\nu(x)}^{v(x)}f(t)d t=f(v(x))-F(u(x))\)              

                                       

Differentiating both sides of this equation with respect to \(x\) gives the equation we want:

\(\frac{d}{d x} \int_{u(x)}^{v(x)} f(t) d t=f(v(x)) \frac{d v}{d x}-f(u(x)) \frac{d u}{d x}\)

\(=F^{\prime}(v(x)) \frac{d v}{d x}-F^{\prime}(u(x)) \frac{d u}{d x}\)

\(=f(v(x)) \frac{d y}{\partial x}-f(u(x)) \frac{d u}{d x}\)        

                           

Use Leibniz's Rule to find the derivatives of the functions in Exercises 31-38.

        \(f(x)=\int_{1 / x}^{x} \frac{1}{t} d t\)

Equation Transcription:

Text Transcription:

f

[a,b]

u(x)

v(x)

x

[a,b]

d/dx integral over u(x) ^v(x) f(t)dt=f(v(x))dv/dx-f(u(x))du/dx

F

f

[a,b]

Integral over u(x) ^v(x) f(t)dt=f(v(x))-F(u(x))

x

d/dx integral over u(x) ^v(x) f(t)dt=d/dx[F(v(x))-F(u(x))]

=F'(v(x)dv/dx-F'(u(x))dudx

=f(v(x))dv/dx-f(u(x))dudx

f(x)= integral over 1/x ^x 1/t dt

ANSWER:

SOLUTION:

Step 1:

In this question using Leibniz’s Rule we have  to find the derivatives of the function.

Add to cart


Study Tools You Might Need

Not The Solution You Need? Search for Your Answer Here:

×

Login

Login or Sign up for access to all of our study tools and educational content!

Forgot password?
Register Now

×

Register

Sign up for access to all content on our site!

Or login if you already have an account

×

Reset password

If you have an active account we’ll send you an e-mail for password recovery

Or login if you have your password back