Find f'(0.5) for f(x)=f on the top "x" on the bottom "0" Arc tan(t)dt
Or, more conventionally (I think)
f(x) = ∫[0,x] arctan(t) dt
By the 2nd FTC,
f'(x) = arctan(x)
so,
f'(0.5) = arctan(0.5) ≈ 0.46
Well, well, well, looks like we have ourselves a mathematical feast! Alright, let's dive into it.
To find f'(0.5), we first need to differentiate the function f(x) = ∫[0 to x] arctan(t) dt with respect to x. Before we do that, let's take a deep breath and remember our differentiation rules.
Now, here comes the fun part! Since we have an integral (the fancy "∫" symbol) involved, we'll have to apply the Fundamental Theorem of Calculus and use the chain rule.
But hey, don't worry, I'll guide you through it. When we differentiate an integral like this, we essentially treat "x" as the upper limit of integration. So, we can rewrite our function as:
f(x) = ∫[0 to x] arctan(t) dt
Now, let's apply the Fundamental Theorem of Calculus. According to this theorem, if we have a function g(x) = ∫[a to x] f(t) dt, then:
g'(x) = f(x)
Hmm, let's see. In our case, we can rewrite f(x) as:
f(x) = ∫[0 to x] arctan(t) dt
Taking the derivative of both sides, we have:
f'(x) = d/dx [∫[0 to x] arctan(t) dt]
Using the chain rule, we differentiate the inside of the integral with respect to x and multiply it by the derivative of the upper limit of integration:
f'(x) = arctan(x) * d/dx (x)
Now, the derivative of x with respect to x is simply 1:
f'(x) = arctan(x) * 1
Simplifying further, we have:
f'(x) = arctan(x)
Almost there! Now, to find f'(0.5), we just need to substitute x = 0.5 into our expression:
f'(0.5) = arctan(0.5)
And there you have it! The value of f'(0.5) is equal to arctan(0.5).
To find the derivative of f(x) = ∫(x to 0) arctan(t) dt, we can apply the Fundamental Theorem of Calculus and the chain rule.
According to the Fundamental Theorem of Calculus, if F(x) is the antiderivative of f(x), then the derivative of ∫(a to x) f(t) dt is F(x).
Let's start by finding the antiderivative of f(x) = arctan(t):
∫ arctan(t) dt = t*arctan(t) - ∫(1+t^2) dt
Simplifying the above integral, we get:
∫ arctan(t) dt = t*arctan(t) - (t + 1/2 * t^3) + C
Now, applying the Fundamental Theorem of Calculus, the antiderivative of f(x) is given by:
F(x) = x*arctan(x) - (x + 1/2 * x^3)
To find f'(0.5), we just need to evaluate the derivative of F(x) at x=0.5.
Therefore, f'(0.5) = d/dx [F(x)] evaluated at x=0.5.
Differentiating F(x) with respect to x, we get:
f'(x) = arctan(x) + x / (1 + x^2) - 1/2 * x^2
Substituting x = 0.5 into f'(x), we have:
f'(0.5) = arctan(0.5) + 0.5 / (1 + 0.5^2) - 1/2 * 0.5^2
Calculating these values:
f'(0.5) = 0.463647609 + 0.5 / (1 + 0.25) - 1/2 * 0.25
f'(0.5) = 0.463647609 + 0.5 / 1.25 - 0.03125
f'(0.5) = 0.463647609 + 0.4 - 0.03125
f'(0.5) = 0.832397609
To find f'(0.5) for the given function f(x) = ∫(x to 0) arctan(t) dt, we need to find the derivative of f(x) and evaluate it at x = 0.5.
Let's begin by calculating the derivative of f(x) using the Fundamental Theorem of Calculus. According to this theorem, if a function f(x) is defined as an integral of another function, then its derivative can be obtained by evaluating the integrand at the upper limit of the integral.
In this case, we have f(x) = ∫(x to 0) arctan(t) dt. To find the derivative, we will differentiate the integral with respect to x and substitute x with 0.5 in the resulting expression.
Differentiating the integral involves applying the Chain Rule, since the upper limit of the integral depends on x. Hence, we have:
f'(x) = d/dx ∫(x to 0) arctan(t) dt.
Applying the Chain Rule, we get:
f'(x) = [d/dx (x)] * arctan(0) - arctan(x).
Since arctan(0) = 0, the equation simplifies to:
f'(x) = -arctan(x).
Now, to find f'(0.5), we substitute x = 0.5 into the derivative expression:
f'(0.5) = -arctan(0.5).
Therefore, f'(0.5) is equal to -arctan(0.5).
Alternatively, you can directly differentiate f(x) = ∫(x to 0) arctan(t) dt using the First Fundamental Theorem of Calculus and then evaluate the resulting expression at x = 0.5. The derivative of f(x) with respect to x can be written as:
f'(x) = arctan(0) - arctan(x).
Substituting x = 0.5, we have:
f'(0.5) = arctan(0) - arctan(0.5),
which gives the same result as above, -arctan(0.5).