How do you find the derivative of #f(x) =intcos 2t dt# over #[x,pi/4]#?

Question

Scarlett Allison · Accepted Answer

So, #f(x) = int_x^(pi/4) cos(2t) dt = - int_(pi/4)^x cos(2t) dt#. The fundamental theorem says that #f'(x) = - cos(2x)#. Theorem (fundamental). Let #f# be a continuous function over an interval #I#.  If #a in I# is a constant, the function defined by
#F(x) = int_a^x f(t) dt #
is differentiable over #I#, and #F'(x) = f(x)# for all #x in I#.

Charles Anctil · Answer

undefined To find the derivative of $ f(x) = \int_{x}^{\frac{\pi}{4}} \cos(2t) \, dt $, we can apply the Fundamental Theorem of Calculus and the Chain Rule. The derivative of the integral function with respect to $ x $ is equal to the integrand evaluated at the upper limit of integration multiplied by the derivative of the upper limit of integration with respect to $ x $, minus the integrand evaluated at the lower limit of integration multiplied by the derivative of the lower limit of integration with respect to $ x $.

Thus, the derivative of $ f(x) $ is:

$$ f'(x) = \cos\left(2 \cdot \frac{\pi}{4}\right) \cdot \frac{d}{dx}\left(\frac{\pi}{4}\right) - \cos(2x) \cdot \frac{d}{dx}(x) $$

$$ f'(x) = \cos\left(\frac{\pi}{2}\right) \cdot 0 - \cos(2x) \cdot 1 $$

$$ f'(x) = -\cos(2x) $$