Solution Found!
Solved: Leibniz’s Rule In applications, we sometimes
Chapter 5, Problem 31AAE(choose chapter or problem)
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.