Question

In Exercises \(1-20,\) find \(d y / d x\). $$y=\int_{\sin x}^{\cos x} t^{2} d t$$

Short answer

The derivative \(dy/dx = [-(\cos x)^2 \cdot \sin(x)] - [(\sin x)^2 \cdot \cos(x)] \

Step-by-step solution

Identify the form

The given function is of the form \(y=\int_{f(x)}^{g(x)} h(t) dt\) which falls under the Leibniz Rule and the chain rule for differentiating the limits will be applied. Here, \(f(x)=\sin(x), g(x)=\cos(x), h(t)=t^2\).

Apply the Leibniz Rule

We apply the first part of Leibniz Rule: if \( y=\int_{a}^{g(x)} h(t) dt \) then \( dy/dx = h(g(x))g'(x) \). So for \( y=\int_{a}^{\cos x} t^{2} dt \) we get that the derivative equals to \((\cos x)^2 \times -\sin(x)\)

Apply the Leibniz Rule again

We apply the second part of Leibniz Rule: if \( y=\int_{f(x)}^{b} h(t) dt \) then \( dy/dx = -h(f(x))f'(x) \). So for \( y=\int_{\sin x}^{b} t^{2} dt \) we get that the derivative equals to \(-( \sin x )^2 \times \cos(x) \)

Add the results

Since our integral has both limits as functions of x, we need to add both derivatives calculated. Thus, \(dy/dx = [ - (\cos x)^2 \cdot \sin(x) ] - [ (\sin x)^2 \cdot \cos(x) ]\)