Differentiation

New Zealand

Level 7 - NCEA Level 2

Lesson

This means using the derivative we can

- find the gradient function, also called the derivative
- evaluate the gradient at a point, say $x=a$
`x`=`a` - find the equation of the tangent at that point, because we have the gradient and a point

Let's work through an example

Find the equation of the tangent to the curve $f(x)=\frac{1}{4}x^2-2x+3$`f`(`x`)=14`x`2−2`x`+3 at the point where $x=4$`x`=4.

**Think**:

This is our plan of attack

a) find the derivative function $f'(x)$`f`′(`x`)

b) find the value of the derivative at the point where $x=4$`x`=4, this will be the gradient of the tangent

c) find the full coordinate on the curve by finding $f(4)$`f`(4)

d) use the point and the gradient to find the equation of the tangent.

a) find the derivative function $f'(x)$`f`′(`x`)

The function is $f(x)=\frac{1}{4}x^2-2x+3$`f`(`x`)=14`x`2−2`x`+3, so using the power rule on each term we find that the derivative is $f'(x)=\frac{x}{2}-2$`f`′(`x`)=`x`2−2

b) find the value of the derivative at the point where $x=4$`x`=4, this will be the gradient of the tangent.

$f'(x)=\frac{x}{2}-2$`f`′(`x`)=`x`2−2, so $f'(4)=\frac{4}{2}-2=0$`f`′(4)=42−2=0

This means that at the point where x=4, the gradient is 0, meaning that the tangent at that point is horizontal.

c) find the full coordinate on the curve by finding f(4)

The function is $f(x)=\frac{1}{4}x^2-2x+3$`f`(`x`)=14`x`2−2`x`+3, so $f(4)=\frac{1}{4}4^2-2\times4+3=4-8+3=-1$`f`(4)=1442−2×4+3=4−8+3=−1. So the coordinate on the curve is $(4,-1)$(4,−1).

d) use the point and the gradient to find the equation of the tangent.

The point is $(4,-1)$(4,−1) and the gradient is $0$0. The equation of the horizontal line is $y=-1$`y`=−1.

By considering the graph of $f\left(x\right)=-6$`f`(`x`)=−6, find $f'\left(4\right)$`f`′(4).

By considering the graph of $f\left(x\right)=2x-3$`f`(`x`)=2`x`−3, find $f'$`f`′$\left(-4\right)$(−4).

Consider the parabola $f\left(x\right)=x^2+3x-10$`f`(`x`)=`x`2+3`x`−10.

Solve for the $x$

`x`-intercepts. Write all solutions on the same line, separated by a comma.Determine the gradient of the tangent at the positive $x$

`x`-intercept.

Consider the tangent to the curve $f\left(x\right)=5\sqrt{x}$`f`(`x`)=5√`x` at the point $\left(\frac{1}{9},\frac{5}{3}\right)$(19,53).

Firstly, find the gradient of the function $f\left(x\right)=5\sqrt{x}$

`f`(`x`)=5√`x`at $x=\frac{1}{9}$`x`=19.Hence find the equation of the tangent to the curve $f\left(x\right)=5\sqrt{x}$

`f`(`x`)=5√`x`at the point $\left(\frac{1}{9},\frac{5}{3}\right)$(19,53).Express the equation of the tangent line in the form $y=mx+c$

`y`=`m``x`+`c`.

Consider the tangent to the curve $f\left(x\right)=\frac{2}{x^3}$`f`(`x`)=2`x`3 at $x=-2$`x`=−2.

Firstly, find the gradient of the function $f\left(x\right)=\frac{2}{x^3}$

`f`(`x`)=2`x`3 at $x=-2$`x`=−2.Determine the $y$

`y`-coordinate of the point on the tangent line.Hence find the equation of the tangent to the curve $f\left(x\right)=\frac{2}{x^3}$

`f`(`x`)=2`x`3 at $x=-2$`x`=−2.Express the equation of the tangent line in the form $y=mx+c$

`y`=`m``x`+`c`.