# 5.1 Differentiation (first principles, rules) and sketching graphs  (Page 2/3)

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## Summary of differentiation rules

Given two functions, $f\left(x\right)$ and $g\left(X\right)$ we know that:

 $\frac{d}{dx}b=0$ $\frac{d}{dx}\left({x}^{n}\right)=n{x}^{n-1}$ $\frac{d}{dx}\left(kf\right)=k\frac{df}{dx}$ $\frac{d}{dx}\left(f+g\right)=\frac{df}{dx}+\frac{dg}{dx}$

## Rules of differentiation

1. Find ${f}^{\text{'}}\left(x\right)$ if $f\left(x\right)=\frac{{x}^{2}-5x+6}{x-2}$ .
2. Find ${f}^{\text{'}}\left(y\right)$ if $f\left(y\right)=\sqrt{y}$ .
3. Find ${f}^{\text{'}}\left(z\right)$ if $f\left(z\right)=\left(z-1\right)\left(z+1\right)$ .
4. Determine $\frac{dy}{dx}$ if $y=\frac{{x}^{3}+2\sqrt{x}-3}{x}$ .
5. Determine the derivative of $y=\sqrt{{x}^{3}}+\frac{1}{3{x}^{3}}$ .

## Applying differentiation to draw graphs

Thus far we have learnt about how to differentiate various functions, but I am sure that you are beginning to ask, What is the point of learning about derivatives? Well, we know one important fact about a derivative: it is a gradient. So, any problems involving the calculations of gradients or rates of change can use derivatives. One simple application is to draw graphs of functions by firstly determine the gradients of straight lines and secondly to determine the turning points of the graph.

## Finding equations of tangents to curves

In "Average Gradient and Gradient at a Point" we saw that finding the gradient of a tangent to a curve is the same as finding the gradient (or slope) of the same curve at the point of the tangent. We also saw that the gradient of a function at a point is just its derivative.

Since we have the gradient of the tangent and the point on the curve through which the tangent passes, we can find the equation of the tangent.

Find the equation of the tangent to the curve $y={x}^{2}$ at the point (1,1) and draw both functions.

1. We are required to determine the equation of the tangent to the curve defined by $y={x}^{2}$ at the point (1,1). The tangent is a straight line and we can find the equation by using derivatives to find the gradient of the straight line. Then we will have the gradient and one point on the line, so we can find the equation using: $y-{y}_{1}=m\left(x-{x}_{1}\right)$ from grade 11 Coordinate Geometry.

2. Using our rules of differentiation we get: ${y}^{\text{'}}=2x$

3. In order to determine the gradient at the point (1,1), we substitute the $x$ -value into the equation for the derivative. So, ${y}^{\text{'}}$ at $x=1$ is: $m=2\left(1\right)=2$

4. $\begin{array}{ccc}\hfill y-{y}_{1}& =& m\left(x-{x}_{1}\right)\hfill \\ \hfill y-1& =& \left(2\right)\left(x-1\right)\hfill \\ \hfill y& =& 2x-2+1\hfill \\ \hfill y& =& 2x-1\hfill \end{array}$
5. The equation of the tangent to the curve defined by $y={x}^{2}$ at the point (1,1) is $y=2x-1$ .

## Curve sketching

Differentiation can be used to sketch the graphs of functions, by helping determine the turning points. We know that if a graph is increasing on an interval and reaches a turning point, then the graph will start decreasing after the turning point. The turning point is also known as a stationary point because the gradient at a turning point is 0. We can then use this information to calculate turning points, by calculating the points at which the derivative of a function is 0.

If $x=a$ is a turning point of $f\left(x\right)$ , then: ${f}^{\text{'}}\left(a\right)=0$ This means that the derivative is 0 at a turning point.

Take the graph of $y={x}^{2}$ as an example. We know that the graph of this function has a turning point at (0,0), but we can use the derivative of the function: ${y}^{\text{'}}=2x$ and set it equal to 0 to find the $x$ -value for which the graph has a turning point.

#### Questions & Answers

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