Derivative of h(x) = ∫(x^3 – 12 + t^2)^(-1/2) dt with respect to x is constant- a comprehensive explanation.

If h(x)=∫x3−12+t2−−−−−√ⅆt for x≥0, then h′(x)=

To find the derivative of the function h(x) = ∫(x^3 – 12 + t^2)^(-1/2) dt with respect to x, we can use the Fundamental Theorem of Calculus along with the Chain Rule

To find the derivative of the function h(x) = ∫(x^3 – 12 + t^2)^(-1/2) dt with respect to x, we can use the Fundamental Theorem of Calculus along with the Chain Rule.

Let’s start by using the Fundamental Theorem of Calculus:

∫(x^3 – 12 + t^2)^(-1/2) dt = F(t) + C,

where F(t) is an antiderivative of (x^3 – 12 + t^2)^(-1/2), and C is the constant of integration.

Now, we need to differentiate both sides of this equation with respect to x:

d/dx [∫(x^3 – 12 + t^2)^(-1/2) dt] = d/dx [F(t) + C].

The derivative on the left side can be written as:

h'(x) = d/dx [F(t)] + d/dx [C].

Now, we focus on the first term on the right side. We’ll use the Chain Rule to differentiate F(t) with respect to x. Since F(t) is an antiderivative of (x^3 – 12 + t^2)^(-1/2), we can write:

dF/dx = dF/dt * dt/dx.

The derivative dF/dt can be obtained by substituting (x^3 – 12 + t^2)^(-1/2) into the integrand of F(t):

dF/dt = (x^3 – 12 + t^2)^(-1/2).

The derivative dt/dx is simply 1 (since t is not explicitly dependent on x):

dt/dx = 1.

Combining these results, we have:

dF/dx = (x^3 – 12 + t^2)^(-1/2).

Now, let’s revisit the right side of the equation. Since dF/dx is the derivative of F(t) with respect to x, and F(t) is a constant with respect to x, the derivative of F(t) with respect to x is zero:

d/dx [F(t)] = 0.

So, the first term on the right side simplifies to zero:

d/dx [F(t)] + d/dx [C] = 0 + d/dx [C] = d/dx [C].

This means that d/dx [∫(x^3 – 12 + t^2)^(-1/2) dt] is simply the derivative of the constant C, which is zero.

Therefore, the derivative of h(x) with respect to x is:

h'(x) = 0.

In summary, h'(x) = 0.

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