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 derivative of a function is defined as:
\[ f^{\prime}(x) = \lim_{x \to 0}\frac{f(x + h) - f(x)}{h} \]
fx <- function(x) {
return (x^3 + 2*x^2)
}Apply the function fx to the derivative formula. An example would look like:
(fx(2 + 0.00000001) - fx(2))/ 0.00000001
# where h = 0.00000001Create the R function to calculate the derivative of fx for any given x value using limits.
## Parameters
# ---------------------------------------
# 1. eq = function to be applied to x
# 2. x = input value to function
# 3. h = limit value approaching 0
deriv_lim <- function(eq, x, h) {
f_xh <- eq(x + h) # f(x + h)
f_x <- eq(x) # f(x)
return ((f_xh - f_x) / h)
}
# create a function to calculate the actual derivative value (analytical form)
f <- function(x) (x^3 + 2*x^2)
deriv_f <- function(x) {}
body(deriv_f) <- D(body(f), 'x')
# derivative of f(x) = x^3 + 2*x^2
body(deriv_f) ## 3 * x^2 + 2 * (2 * x)# example : calculate derivative of x for 1 : 3
deriv_f(1:3)## [1] 7 20 39n <- 1:20
results = data.frame(x = n,
deriv_limit = deriv_lim(fx, n, 0.00000001),
deriv_calc = deriv_f(n))Results:
## x deriv_limit deriv_calc
## 1 7.0000 7
## 2 20.0000 20
## 3 39.0000 39
## 4 64.0000 64
## 5 95.0000 95
## 6 132.0000 132
## 7 175.0000 175
## 8 224.0000 224
## 9 279.0000 279
## 10 340.0000 340
## 11 407.0000 407
## 12 480.0001 480
## 13 559.0000 559
## 14 644.0001 644
## 15 735.0000 735
## 16 832.0001 832
## 17 935.0001 935
## 18 1044.0001 1044
## 19 1159.0001 1159
## 20 1280.0001 1280Now, 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 \(\Delta x\) value (say 1e-6) and then compute the approximation to the area under the curve.
f2 <- function(x) {
return (3*x^2 + 4*x)
}
# create the range of small intervals using a delta of 1e-6
intervals <- seq(from=1, to=3, by=1e-6)
head(intervals, 10)## [1] 1.000000 1.000001 1.000002 1.000003 1.000004 1.000005 1.000006
## [8] 1.000007 1.000008 1.000009Number of intervals = 2000001
area.curve <- function(func, x, delta) {
# calculate the heights of each interval
height <- func(x)
# width will be the delta value
width <- delta
return (sum( height * width))
}
(area_approx <- area.curve(func=f2, x=intervals, delta=1e-6))## [1] 42.00002Let’s compare this value calculated by R using the integrate function.
(area_R <- integrate(f2, 1, 3))## 42 with absolute error < 4.7e-13R Integrate function = 42
Approximataion function = 42.000023
Let: \[u = f(x) \] \[du = f^{\prime}(x) dx\] \[dv = g^{\prime}(x) dx\] \[v = g(x) \]
Then the Integration by Parts formula becomes:
\[ \int{u dv} = uv - \int{v du} \]
Use integration by parts to solve for \(\int{sin(x)cos(x)}dx\)
For this problem, set the following:
\(u = -cos(x)\)
\(du = sin(x)dx\)
\(dv = -sin(x)dx\)
\(v = cos(x)\)
Apply the Integration by Parts formula:
\[\int{sin(x)cos(x)}dx = -cos(x) cos(x) - \int{ -cos(x) -sin(x) dx }\]
re-write to:
\[\int{sin(x)cos(x)}dx = -cos(x) cos(x) - \int{cos(x) sin(x) dx }\]
adding \(\int{cos(x) sin(x) dx}\) to LHS we get:
\[ 2 * \int{cos(x) sin(x) dx} = -cos^2(x) \]
Finally
\[\int{cos(x) sin(x) dx} = -\frac{cos^2(x)}{2} + C \]
Use integration by parts to solve for \(\int{x^2 e^x}\)
For this problem, set the following:
\(u = x^2\)
\(du = 2xdx\)
\(dv = e^xdx\)
\(v = e^x\)
u = x^2 v = e^x du = 2xdx dv = e^xdx
\[\int{x^2 e^x} = x^2 e^x - \int{e^x 2xdx}\]
\[ = x^2 e^x - 2 \int{x e^x dx}\]
Calculate second anti-derivative for
\[ \int{x e^x dx}\]
let:
\(u = x\)
\(du = 1\)
\(v = e^x\)
\(dv = e^x\)
\[ = x e^x - \int{ 1 e^x dx} = x e^x - e^x \]
Putting it all together:
\[ = x^2 e^x - 2(x e^x - e^x) + C = x^2 e^x -2xe^x + 2e^x + C\]
What is \(\frac{d}{dx} (x cos(x))\)?
To solve use the Product Rule which states that:
\[(fg)^{\prime} = f^{\prime} * g + f * g^{\prime} \]
\(f = x\) \(f^{\prime} = 1\)
\(g = cos(x)\) \(g^{\prime} = -sin(x)\)
\(f^{\prime} * g + f * g^{\prime} = 1*cos(x) + x(-sin(x)) = cos(x) - x(sin(x))\)
What is \(\frac{d}{dx}(e^x)^4\)?
Use the Power Rule and regroup: \(\frac{d}{dx}(e^4x)\)
Next apply the Chain Rule where u = 4x
\[(e^4x) * \frac{d}{dx}4x = 4 \cdot \frac{d}{dx}x \cdot (e^4x)\]
\[ = 4 \cdot 1(e^4x) = 4e^4x\]