Assignment: Simplify the derivative (with respect to x) of an integral (with respect to t), without calculating the integral.

[I'm here because idk]

So I have been doing old exams for a calculus course and I encountered the problem in the attached image.
I have no clue how to approach it without calculating the integral first. I hope someone here does. Thanks in advance.

 

[Dr.Peterson]

So I have been doing old exams for a calculus course and I encountered the problem in the attached image.
I have no clue how to approach it without calculating the integral first. I hope someone here does. Thanks in advance.

Have you learned the Fundamental Theorem of Calculus? Here is an explanation, in which the third example is this type.

Here are some more interesting examples, more like yours.

 

[pka]

So I have been doing old exams for a calculus course and I encountered the problem in the attached image.
I have no clue how to approach it without calculating the integral first. I hope someone here does. Thanks in advance.

Here is a template $\displaystyle{D_x}\int_{{x^2}}^{{x^3}} {\sqrt {1 + \log (t)} dt = 3{x^2}\sqrt {1 + \log ({x^3})} - 2x\sqrt {1 + \log ({x^2})} }$

 

[skeeter]

for $u \text{ and } v$ both functions of $x$

$\displaystyle \dfrac{d}{dx} \bigg[\int_u^v f(t) \, dt \bigg] = f(v) \cdot \dfrac{dv}{dt} - f(u) \cdot \dfrac{du}{dt}$

 

[blamocur]

for $u \text{ and } v$ both functions of $x$

$\displaystyle \dfrac{d}{dx} \bigg[\int_u^v f(t) \, dt \bigg] = f(v) \cdot \dfrac{dv}{dt} - f(u) \cdot \dfrac{du}{dt}$

dv/dx, du/dx ?

 

[Steven G]

dv/dx, du/dx ?

Yep. I missed that one. Good catch!

 

[skeeter]

dv/dx, du/dx ?

sorry … “t” on the brain