Saturday, February 20, 2016

VOLUME OF HEXAGONAL PRISM

DIFFERENTIATION OF EXPONENTIAL FUNCTIONS

DERIVATIVE OF THE EXPONENTIAL FUNCTION

The derivative of e^x is quite remarkable. The expression for the derivative is the same as the expression that we started with; that is, e^x!

$What does this mean?$ It means the slope is the same as the function value (the y-value) for all points on the graph.

Example: Let's take the example when x = 2. At this point, the y-value is e² ≈ 7.39.
Since the derivative of e^x is e^x, then the slope of the tangent line at x = 2 is also e² ≈ 7.39.
We can see that it is true on the graph:

Let's now see if it is true at some other values of x.

We can see that at x = 4, the y-value is 54.6 and the slope of the tangent (in red) is also 54.6.
At x = 5, the y-value is 148.4, as is the value of the derivative and the slope of the tangent (in green).

DIFFERENTIATION OF EXPONENTIAL FUNCTIONS

The derivative of f(x) = b^x is given by

f '(x) = b^x ln b

Note: if f(x) = e^x , then f '(x) = e^x

Example 1: Find the derivative of f(x) = 2^x

Solution to Example 1:

· Apply the formula above to obtain

f '(x) = 2^x ln 2

Example 2: Find the derivative of f(x) = 3^x + 3x²

Solution to Example 2:

· Let g(x) = 3^x and h(x) = 3x², function f is the sum of functions g and h: f(x) = g(x) + h(x). Use the sum rule, f '(x) = g '(x) + h '(x), to find the derivative of function f

f '(x) = 3^x ln 3 + 6x

Example 3: Find the derivative of f(x) = e^x / ( 1 + x )

Solution to Example 3:

· Let g(x) = e^x and h(x) = 1 + x, function f is the quotient of functions g and h: f(x) = g(x) / h(x). Hence we use the quotient rule, f '(x) = [ h(x) g '(x) - g(x) h '(x) ] / h(x)², to find the derivative of function f.

g '(x) = e^x

h '(x) = 1

f '(x) = [ h(x) g '(x) - g(x) h '(x) ] / h(x)²

= [ (1 + x)(e^x) - (e^x)(1) ] / (1 + x)²

Multiply factors in the numerator and simplify

f '(x) = x e^x / (1 + x)²

Example 4: Find the derivative of f(x) = e^{2x + 1}

Solution to Example 4:

· Let u = 2x + 1 and y = e^u, Use the chain rule to find the derivative of function f as follows.

f '(x) = (dy / du) (du / dx)

· dy / du = e^u and du / dx = 2

f '(x) = (e^u)(2) = 2 e^u

· Substitute u = 2x + 1 in f '(x) above

f '(x) = 2 e^{2x + 1}

Exercises Find the derivative of each function.

1 - f(x) = e^x 2^x

2 - g(x) = 3^x - 3x³

3 - h(x) = e^x / (2x - 3)

4 - j(x) = e^{(x2 + 2)}

solutions to the above exercises

1 - f '(x) = e^x 2^x ( ln 2 + 1)

2 - g '(x) = 3^x ln 3 - 9x²

3 - h '(x) = e^x(2x -5) / (2x - 3)²

4 - j '(x) = 2x e^{(x2 + 2)}

Solve the following problem

Differentiate y=e^(ax+b) where a and b are constants

Let u = ax + b and y = e ^u

Use the chain rule to find the derivative of function as follows

dy/dx = (dy / du) (du / dx)

dy / du = e ^u and du / dx = a ( Since the derivative of the constant b is 0)

Hence dy/dx = (e ^u)(a) = a e ^u

Substitute u = ax + b in dy/dx above

dy/dx = a e ^ax + b

VOLUME OF A TRIANGULAR PRISM

SUMMARY

B stands for the area of the base
Since our bases are triangles, we can use (1/2)bh for B
We can call the height of the triangle h₁ and the height of the prism h₂ to avoid confusion
Volume is measured in units cubed, so we have 50 cm³

NOTES

1. The volume of a prism is V=Bh
2. B is the area of the base of the prism
3. h is the height of the prism
4. The base of our prism is a triangle, so we can use (1/2)bh, the area of a triangle, for B
5. We can call the height of the triangle h₁ and the height of the prism h₂ to avoid confusion
1. V is the volume, which is what we are trying to find
2. b is the base of the triangle, which is 5 cm
3. h₁ is the height of the triangle, which is 2 cm
4. h₂ is the height of the prism, which is 10 cm
1. We can plug 5 in for 'b', 2 in for 'h₁', and 10 in for 'h₂' in our equation
1. We need to use the order of operations to simplify the right hand side
2. First multiply 5•2 in the innermost parentheses
3. Then multiply (1/2)•10 within the next set of parentheses
4. This is the same as 10/2, or 5
5. Then multiply 5•10 to get 50 cm³