Use the Fundamental Theorem of Calculus to find G'(x) if:
G(x)=∫ up(x²) bottom(1) cost dt
this is just the chain rule in reverse.
G'(x) = cos(x^2) (2x)
To find the derivative of G(x) using the Fundamental Theorem of Calculus, we first need to re-express G(x) as a definite integral.
The Fundamental Theorem of Calculus states that if F(x) is an antiderivative of a function f(x) on an interval [a, x], then
∫ from a to x f(t) dt = F(x) - F(a)
In our case, G(x) is given by:
G(x) = ∫ from 1 to x p(t²)cos(t) dt
To find the derivative G'(x), we need to differentiate G(x) with respect to x. However, since G(x) is an integral involving a variable upper limit, we need to apply the Leibniz Rule for differentiating under the integral sign.
The Leibniz Rule states that if the function f(x, t) is continuous on a region R defined by a ≤ x ≤ b and α ≤ t ≤ β, and if both the function f(x, t) and its partial derivative ∂f/∂x are continuous on R, then the function:
G(x) = ∫ from α to β f(x, t) dt
is differentiable for a ≤ x ≤ b, and its derivative is given by:
G'(x) = d/dx ∫ from α to β f(x, t) dt = ∫ from α to β (∂f/∂x) dx
In our case, we have:
G(x) = ∫ from 1 to x p(t²)cos(t) dt
Let's differentiate this integral with respect to x using the Leibniz Rule.
G'(x) = d/dx ∫ from 1 to x p(t²)cos(t) dt
By applying the Leibniz Rule, we can interchange the derivative and the integral sign:
G'(x) = ∫ from 1 to x (∂/∂x)(p(t²)cos(t)) dt
Now, we need to differentiate p(t²)cos(t) with respect to x. However, we only know the expression for p(t), not p(t²). Assuming p(t) is differentiable with respect to t, we can use the chain rule to differentiate p(t²) with respect to x.
Let's denote u = t², then du/dt = 2t, and we can rewrite p(t²) as p(u). Applying the chain rule, we get:
d/dx p(t²) = d/dx p(u) = dp/du * du/dx
Since t = √u, we have du/dx = d/dx (√u) = (1/2√u)(2t) = t/√u
Therefore,
d/dx p(t²) = dp/du * du/dx = dp/du * (t/√u)
Now, we can substitute this result back into G'(x):
G'(x) = ∫ from 1 to x (∂/∂x)(p(t²)cos(t)) dt
G'(x) = ∫ from 1 to x (∂p/∂u * (t/√(t²)) * cos(t)) dt
Simplifying further, we have:
G'(x) = ∫ from 1 to x (∂p/∂u * (t/t) * cos(t)) dt
G'(x) = ∫ from 1 to x (∂p/∂u * cos(t)) dt
Therefore, G'(x) = ∫ from 1 to x (∂p/∂u * cos(t)) dt