Differentiation evaluates the rate of change of a function, which is the same as the gradient on a graph. A gradient on a graph is given as `frac(text(change in )y)(text(change in )x)`. A differentiation is shown as `frac(dy)(dx)` - often said as differentiate dy by dx.
The process is to take a term `ax^n`
>multiply the coefficient by the power<
>subtract 1 from the power<
Note that if there is a constant in the expression, then that constant is dropped.
If `y = ax^n`, then `frac(dy)(dx) = anx^(n-1)`.
If a function `f(x)` is differentiated, it is shown as `f`(x)`. A function `f(x) = ax^n` would differentiate to `f`(x) = anx^(n-1)`.
The power, `n`, can be positive, negative, a fraction or a decimal.
A curve has an equation of `y = 5x^3 - 7x^2 + 3x -5`. Find `frac(dy)(dx)`.
For each term, multiply the coefficient by the power, and subtract 1 from the power.
`5x^3` becomes `3 xx 5x^(3-1) = 15x^2`
`-7x^2` becomes `2 xx -7x^(2-1) = -14x`
`3x` becomes `1 xx 3x^(1-1) = 3`
The contant -5 is dropped
`frac(dy)(dx) = 15x^2 -14x + 3`
Answer: `15x^2 - 14x + 3`
A curve has the equation `y=-2x^3+4x-7`. A tangent is drawn to the curve when `x=-2`, and another tangent is drawn to the curve when `x=2`. Describe the relationship between these two tangents.
The gradient can be found by differentiating the equation.
`frac(dy)(dx) = 3 xx -2x^(3 - 1) + 1 xx 4x^(1-1) = -6x^2 + 4`
Substitute the values of `x` from the coordinates into the equation.
Both gradients are the same, therefore the lines are parallel.
Answer: They are parallel.
Reason: `frac(dy)(dx) = -6x^2 +4`.
When `x=-2, frac(dy)(dx) = -6(-2)^2 + 4 = -20`
When `x=2, frac(dy)(dx) = -6(2)^2 + 4 =-20`
Both gradients are the same.