This is a GRE prep question:What's the derivative of f(x)=∫x0cos⁡xttdt

First Foundamental Theorem of Calculus : ddx∫α(x)β(x)f(x,y)dy=∫α(x)β(x)df(x,y)dxdy+dβ(x)dxf(x,β(x))−dα(x)dxf(x,α(x)) where: α(x)=0,β(x)=x,f(x,y)=cos⁡xyy

Note that not only the domain of integration but also the integrand depend on x. Let's (for a moment) write F(X)=∫x0f(x,t)dt=−∫0xf(x,t)dt for your integral. To handle the "double" x-dependence, we write F as the composition of △:R→R2,x→(x,x)F′(x)=△Φ(△(x))⋅△′(x)by the chain rule. Now △′(x)=(1,1) andd1Φ(x1,x2)=−f(x2,x1)d2Φ(x1,x2)=−∫0x1dx2f(x2,t)dt(for the first derivative we used the fundamental theorem). Plugin everything together, we obtainF′(x)=−f(x,x)−∫0xdxf(x,t)dt

First Foundamental Theorem of Calculus : ddx∫α(x)β(x)f(x,y)dy=∫α(x)β(x)df(x,y)dxdy+dβ(x)dxf(x,β(x))−dα(x)dxf(x,α(x)) where: α(x)=0,β(x)=x,f(x,y)=cos⁡xyy

Note that not only the domain of integration but also the integrand depend on x. Let's (for a moment) write F(X)=∫x0f(x,t)dt=−∫0xf(x,t)dt for your integral. To handle the "double" x-dependence, we write F as the composition of △:R→R2,x→(x,x)F′(x)=△Φ(△(x))⋅△′(x)by the chain rule. Now △′(x)=(1,1) andd1Φ(x1,x2)=−f(x2,x1)d2Φ(x1,x2)=−∫0x1dx2f(x2,t)dt(for the first derivative we used the fundamental theorem). Plugin everything together, we obtainF′(x)=−f(x,x)−∫0xdxf(x,t)dt

Expert answers to "This is a GRE prep question:What's the derivative..."

chanyingsauu7

Answered question

2021-11-21

This is a GRE prep question:
What's the derivative of

f (x) = \int_{x}^{0} \frac{\cos x t}{t} d t

Answer & Explanation

Alrew1959

Beginner2021-11-22Added 16 answers

First Foundamental Theorem of Calculus :

\frac{d}{d x} \int_{α (x)}^{β (x)} f (x, y) d y = \int_{α (x)}^{β (x)} \frac{d f (x, y)}{d x} d y + \frac{d β (x)}{d x} f (x, β (x)) - \frac{d α (x)}{d x} f (x, α (x))

where:

α (x) = 0, β (x) = x, f (x, y) = \frac{\cos x y}{y}

William Yazzie

Beginner2021-11-23Added 20 answers

Note that not only the domain of integration but also the integrand depend on x. Let's (for a moment) write $F (X) = \int_{x}^{0} f (x, t) d t = - \int_{0}^{x} f (x, t) d t$ for your integral. To handle the "double" x-dependence, we write F as the composition of $△ : R \to R^{2}, x \to (x, x)$
$F^{'} (x) = △ Φ (△ (x)) \cdot △^{'} (x)$
by the chain rule. Now $△^{'} (x) = (1, 1)$ and
$d_{1} Φ (x_{1}, x_{2}) = - f (x_{2}, x_{1})$
$d_{2} Φ (x_{1}, x_{2}) = - \int_{0}^{x_{1}} d_{x_{2}} f (x_{2}, t) d t$
(for the first derivative we used the fundamental theorem). Plugin everything together, we obtain
$F^{'} (x) = - f (x, x) - \int_{0}^{x} d_{x} f (x, t) d t$

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