- Here we like to find out the derivative for sin(x), cos(x), tan(x), e
^{x}and ln (x). We will start from the first principles. - Recall that differentiation from the first principles is
f ' (x) = __f(x + h) -f(x)__

h**Exploration 1:**

The aim here is to find the limit of [ sin(x+h)-sin(x) ]/h as h tends to zero. We will start a plot of the process with h=0.001.- Make sure your calculator is on Radian mode.
- Set your window: Xmin=0, Xmax = 2p, Xscl=0.5p, Ymin=-1.25, Ymax= 1.25 & Yscl=1
- Press [Y=]. At Y1=, enter (sin(x+0.001)-sin(x))/0.001. Note that h is allowed to be 0.001. You could choose to have a smaller h.
- Plot the graph. Observe your graph. What is the limit of our process [ sin(x+h)-sin(x) ]/h as h tends to zero?
- Enter your guess into Y2 and plot. How close is the process to your guess? Look at the table by pressing [2 nd] [graph]. Compare the values in Y1 and Y2.

Exploration 2:

We can now investigate the limit of [ cos(x+h)-cos(x) ]/h as h tends to zero by simply returning to [Y=] and press.- Go to Y1 and place your cursor in on "s" of sin and press [Del]. This will delete sin.
- Press [2 nd] [Del] to insert.
- Press [cos] and move on to the next "s" of sin and [Del]. Insert cos.
- Repeat (iv) and (v) in Exploration 1.

^{x}and ln (x).Function Derivative sin x cos x cos x -sin x tan x __1__

cos^{2}xe ^{x}e ^{x}ln x __1__

x **Example 1.**Take the derivative of the following functions with respect to x:

- y = sin x + cos x
- y = 2 sin x + x
- g = 1/2sin x- 5 cos x
- g = 2e
^{x}- 3/2 cos x - y = ln x
^{7} - y = ln(2x
^{3}) - y = ln ( 1/√(x) )
- p = log x
- p = log
_{b}x

**Solutions:**- y ' (x) = cos x - sin x
- y ' (x) = 2cos x + 1
- g ' (x) = 1/2cos x + 5 sin x
- g ' (x) = 2e
^{x}+ 3/2 sin x - y = 7 ln x

y '(x) = 7/x - y = ln 2 + ln x
^{3}

y = ln 2 + 3 ln x

dy/dx = 3/x - y = ln 1 - ln √(x)

y = ln 1 - ln (x^{1/2})

y = 0 - 1/2ln x

dy/dx = -1/(2x) - p = log
_{10}x. Here we need to change everything to base e.**Note that log**_{e}x = ln x

p =ln x/ln 10

p = (1/ln 10)ln x

p ' (x) = (1/ln 10)(1/x)

p ' (x) = 1/(xln 10) - p = ln x/ln b p '(x) = 1/(xln b) as in previous solution.

**Example 2.**Find the gradient of the curve at the given value of x.

- y = ln x at x=2
- y = sin x + ln x at x = θ/2
- y = x
^{2}+ ln (3x^{5}) at x= 1

**Solutions:**- y '(x) = 1/x

y ' (2) = 1/2 - y ' (x) = cos x + 1/x

y ' (θ/2) = cos(θ/2) + 1/(θ/2) ; where θ is some fixed angle.

y ' (θ/2) = cos(θ/2) + 2/θ - y = x
^{2}+ ln 3 + ln x^{5}

y = x^{2}+ ln 3 + 5ln x

y ' (x) = 2x + 5/x

y ' (1) = 2(1) + 5/1

y ' (1) = 7.

**Example 3.**Find the equation of the tangent to the curve y = x + ln x at x=1.

**Solution:**

Let us first find the point on the curve. We need to have the y-coordinate. We obtain this by subsituting x=1 into y =x + ln x .

y(1) = 1 + ln 1 = 1Now let us find the derivative of y.

y ' (x) = 1 + 1/x.

y ' (1) = 1 + 1/1

y ' (1) = 2. The gradient of the tangent to the curve at x=1 is 2.Now we will obtain the equation of the tangent at (1,1).

y - 1 = 2 (x -1 )

y = 2x -2 +1

y = 2x - 1.-
**Example 4.**Find the smallest positive value of θ for which y = 2θ + 3cosθ has a gradient of 1/2.

**Solution:**

dy/dθ = 2 -3sinθ

2 -3sinθ = 1/2

2 -1/2 = 3sinθ

3/2 = 3sinθ

1/2 = sinθ

θ = π/6