Write a program to compute the derivative of f(x) = x^3 + 2x^2 at any value of x. Your function should take in a value of x and return back an approximation to the derivative of f(x) evaluated at that value. You should not use the analytical form of the derivative to compute it. Instead, you should compute this approximation using limits.

The limit definition of the derivative: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \]

To calculate this limit we will compute the following two limits and average those results together:

\[ f'(x) = \lim_{h \to 0^+} \frac{f(x+h)-f(x)}{h} \ f'(x) = \lim_{h \to 0^-} \frac{f(x+h)-f(x)}{h} \]

Let \(h\) be a small value, say 1e-6. Once we have the above values we can then average the two limits to get a linear approximation of the actual value of the limit.

derivative <- function(x, h=1e-6){
    # Approximates the derivative of the function f(x)=x^3 + 2x^2 using an approximation of the limit.
    #
    # Inputs:
    #   x = The value for which we need to calculate the derviative.
    #   h = The small value that is added or subtracted from x to approximate the limit.
    #
    # Returns:
    #   The approximate value of the derivative calculated by approximating the limit.

    fxhPos <- (x+h)^3 + 2*(x+h)^2
    fxhNeg <- (x-h)^3 + 2*(x-h)^2
    fx     <- x^3 + 2*x^2
    upperdx <- (fxhPos - fx)/h
    lowerdx <- (fxhNeg - fx)/-h
    dx <- (lowerdx + upperdx)/2
    return(dx)
}

Client Calls:

derivative(4)
## [1] 64
derivative(-2)
## [1] 4

To cross check our results lets calculate these manually: \[ f'(x) = 3x^2 + 4x \] , when \(x = 5\) and \(x = -2\), we expect to get \(f'(4) = 64\) and \(f'(x) = 4\) respectively.

write a program to compute the area under the curve for the function 3x^2+4x in the range x = [1, 3]. You should first split the range into many small intervals using some really small ∆x value (say 1e-6) and then compute the approximation to the area under the curve.

we will be calculating the integral of a given function by taking a series of small intervals and adding up the area of each rectangle. This can be formulated as: \[ \int_{a}^{b} f(x) = \lim_{x \to \infty} \sum_{i=1}^{n} {f(x_i)} \Delta x \]

For this problem we approximate the integral of \(f(x) = 3x^2 + 4x\) over the interval \(x = [1,3]\).

To calculate this value we will divide the interval into a number of small subintervals using \(\Delta x =\) 1e-6. We will then sum up the area of the small rectangles that are created by \(f(x_i) \Delta x\) and return this value.

areaunderCurve <- function(){
    # Compute the area under the curve of 3x^2 + 4x, in the range x = [1, 3]
    #
    # Inputs:
    #   none
    #
    # Returns:
    #   area = The approximation of the area under the curve.
    a <- 1
    b <- 3
    deltax <- 1e-6
    subint <- (b-a)/deltax # here, we are dividing the region into subintervals
    area <- 0
    for(i in 1:subint){
        area <- area + (3*a^2 + 4*a)*deltax # using the above formula
        
        #increment a now by deltax
        a <- a + deltax
    }
    return(area)
}

Client Call:

areaunderCurve()
## [1] 42

Lets cross verify against the known value of the integral: \[ \int_{1}^{3} {3x^2 + 4x} dx = \left. {x^3 + 2x^2} \right|_{1}^{3} = 42 \]

Please solve these problems analytically (i.e. by working out the math) and submit your answers.

Problem 1

Use integration by parts to find \(\int \sin(x) \cos(x) dx\)

From the integration by parts formula \[ \int f g' dx = f g - \int f' g dx \], first we need to pick our \(f\) and \(g'\) and calculate our \(f'\) and \(g\): \[ f = \cos(x) \; f' = -\sin(x) \ g' = \sin(x) \; g = -\cos(x) + C \]

Applying integration by parts we get: \[ \int \sin(x) \cos(x) dx = \cos(x) * -\cos(x) + C - \int -\sin(x) * - \cos(x) dx \ \int \sin(x) \cos(x) dx = -\cos^2(x) + C - \int \sin(x) \cos(x) dx \ 2 \int \sin(x) \cos(x) dx = -\cos^2(x) + C \ \int \sin(x) \cos(x) dx = -\frac{\cos^2(x)}{2} + C \]

Problem 2

Use integration by parts to find \(\int x^2 e^x dx\)

Similar to prior problem, first we need to pick our \(f\) and \(g'\) and calculate our \(f'\) and \(g\): \[ f = x^2 \; f' = 2x \ g' = e^x \; g = e^x + C \]

Applying integration by parts we get: \[ \int x^2 e^x dx = x^2 e^x + C - \int 2x e^x dx \]

Apply integration by parts again to calculate the integral. So picking new values for \(f\) and \(g'\) we get the following: \[ f = 2x \; f' = 2 \ g' = e^x \; g = e^x + C \]

Applying this to the previous interval gives us the following: \[ \int x^2 e^x dx = x^2 e^x + C - (2x e^x - 2 \int e^x dx) \ \int x^2 e^x dx = x^2 e^x - 2x e^x - 2e^x + C \ \int x^2 e^x dx = e^x(x^2 - 2x - 2) + C \]

Problem 3

Find \(\frac{d}{dx} (x \cos(x))\)

Apply the product rule:

\[(\frac{d}{dx}(x \cos(x)) = \cos(x) - x\sin(x))\]

Problem 4

Find \(\frac{d}{dx} (e^{x^4})\)

Apply the chain rule.

\[(\frac{d}{dx}(e^{x^4})=4x^3e^{x^4})\]