Derivative Rules , Common Functions , Tricks , Examples , Exercises ,
Definition:
From first principles is also known as differentiating a function by starting from scratch (Zero) and use algebra to find a general expression for the slope of a curve, at any x value. First principles use the "delta method" Δx (change in x) and Δy (change in y).
Remember:
The Slope is defined as: | m = |
y2 − y1
x2 − x1
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= |
Δy
Δx
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= |
f(x + Δx) − f(x)
Δx
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= |
change in y
change in y
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Putting together, the Slope of a tangent can be written as follow:
dy
dx
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= | lim Δx → 0 |
Δy
Δx
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= | lim Δx → 0 |
f(x + Δx) − f(x)
Δx
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With y = f(x), the derivative y' = f(x)' = |
dy
dx
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Example:
- Find y' from first principles if y = x2 + 4x
- Find the slope of the tangent where x = 1 and also where x = − 6.
- Sketch the curve and both tangents.
Answer:
Assuming f(x) = x2 + 4xNote: y' means "the first derivative", it can also be written dydx
f(x + Δx) = (x + Δx)² + 4(x + Δx)
= x² + 2·xΔx + Δx² + 4x + 4Δx
Therefore
dydx= lim
Δx → 0(x² + 2·xΔx + Δx² + 4x + 4Δx) − (x² + 4x)Δx= lim
Δx → 0(2x + Δx + 4) = 2x + 4.
From first principles y' = 2x + 4.
When x = 1, slope m = 2(1) + 4 = 6.
When x = −6, slope m = 2(-6) + 4 = − 8.- Sketch:
Some tricks!!! |
Find y' from first principles if y = x² + 4x
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To find the derivative of the function y = f(x): do it separately as is an addition of two functions.
For x² (all multiply!) :
- take the power of x down;
- derivate x;
- and reduce the power of x. it gives you: 2 · (x)' · x2 − 1 | (x)' = 1
- 4 ist a constant and can be derivated;
- the derivative of x, (x)' = 1 ⇒ (4x)' = 4
⇒ (x²)' = 2x
For 4x :
Now add both: y' = 2x + 4, and that's the derivative of the function y = f(x).
Rules | Function | Derivative | ||
Multiplication by constant | c·u | c·u’ | ||
Sum Rule | u + v | u’ + v’ | ||
Difference Rule | u − v | u’ − v’ | ||
Power Rule | xn | n·xn − 1 | ||
Product Rule | u · v | u · v’ + u’· v | ||
Quotient Rule |
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Reciprocal Rule |
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Chain Rule (in a different form) |
u(v(x)) | u’(v(x))v’(x) | ||
Chain Rule (as "Composition of Functions") |
u○v | (u’○v)·v' |
Common Functions | Function | Derivative | |||
Constant | c | 0 | |||
Straight Function Linear Function |
x | 1 | |||
Square Function | x² | 2·x | |||
Square Root | √x | ¹/2 · x-1/2 | |||
Exponential | ex | ex | |||
Logarithms | ax ln(x) loga(x) |
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Trigonometry (x is in radians) |
sin(x) cos(x) tan(x) sin-1(x) cos-1(x) tan-1(x) |
cos(x) −sin(x) 1/cos²x -cos(x)/sin²x sin(x)/cos²x 1/1 + x² |
Examples |
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Answer: y' = 2 · 2·x2-1 + 3 = 4x + 3.Find y = dydxfrom first principles if y = 2x² + 3x. - What is the derivative of 5z² + y3 − 7x4, from first principle?
Answer: y' = 5·2·z2-1 + 3·y3-1 − 7·4·x4-1 = 10z + 3y² − 28x3 -
What is ddx((x − 1)(x + 3))?
Answer: 2 methods are possible!
- Method 1: We can use the product rule assuming that:
- Method 2: First expand the terms by multiplying out (remove the parentheses) and then use the Sum or Difference Rule to differentiate it separately
u = x − 1 ==> u' = 1
v = x + 3 ==> v' = 1
Cross multiply and add them as followed:
u·v = u'·v + u·v' = 1·(x + 3) + (x − 1)·1
= x + 3 + x − 1
= 2x + 2 = 2(x + 1)
(x − 1)(x + 3) = x² + 3x − x − 3 = x² + 2x − 3 * ddxx² = 2x * ddx(2x) = 2
Hence, (x − 1)(x + 3) = 2x + 2 = 2(x + 1)* ddx(− 3) = 0
Exercises |
Differentiate from first principles:
i. | y = x² − 4x | ||||
ii. |
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iii. |
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