Question

In Exercises \(1-20,\) find \(d y / d x\). $$y=\int_{x^{3}}^{5} \frac{\cos t}{t^{2}+2} d t$$

Short answer

The derivative of the given function is \(y' = -\frac{3x^2 \cos x^3}{(x^3)^2+2}\)

Step-by-step solution

Understand the Leibniz Rule

Leibniz rule, also known as the Fundamental Theorem of Calculus, provides a way to differentiate functions that are written as integrals with variable limits of integration. The rule is as follows: If we have \(y=\int_{g(x)}^{h(x)} f(t) d t\), then its derivative is given by: \(y' = f[h(x)]*h'(x)-f[g(x)]*g'(x)\). We have to apply this rule to differentiate the given function: \(y=\int_{x^{3}}^{5} \frac{\cos t}{t^{2}+2} d t\)

Differentiate the Inner Function

According to the rule, we first need to differentiate the inner function of the integral \(f(t)=\frac{\cos t}{t^{2}+2}\) with respect to its limits. For \(t=x^3\): The derivative of \(x^3\) is \(3x^2\). So, \(f[g(x)]*g'(x)\) becomes \(f[x^3]*3x^2=\frac{\cos x^3}{(x^3)^2+2}*3x^2\).For \(t=5\): The derivative of the constant \(5\) is \(0\). Therefore, \(f[h(x)]*h'(x)\) equals \(0\) because any number multiplied by \(0\) is \(0\).

Substitute the Values

Substituting the values from steps 1 and 2 into the Leibniz rule: We get: \(y' =0 -\frac{\cos x^3}{(x^3)^2+2}*3x^2\). This simplifies to: \(y' = -\frac{3x^2 \cos x^3}{(x^3)^2+2}\)