## D2 Gradients, tangents and derivatives

A tangent is a line that touches a curve at only one point. Where that point sits along the function curve, determines the slope (i.e. the gradient) of the tangent to that point.

A derivative of a function gives you the gradient of a tangent at a certain point on a curve. If you plug the x value into the derivative function, you will get the slope of the tangent at that point, defined by the x value. Practice evaluating the gradients of these tangents to a curve.

(See also Functions and graphs)

### Gradient of a Curve

In this module we are concerned with finding a formula for the slope or gradient of the tangent at any point on a given curve $$y$$=$$f(x)$$.

The formula $$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$ may be used to find the gradient of a line when two points on the line, $$\left(x_{1},y_{1}\right)$$ and $$\left(x_{2},y_{2}\right)$$ are known.1 There are two special cases that have to be dealt with: horizontal and vertical lines. A horizontal line parallel to the $$x$$-axis with equation of the form $$y=k$$ where $$k$$ is a constant, has a gradient of zero. As a line becomes closer to vertical its gradient gets larger. A vertical line parallel to the $$y$$-axis with equation of the form $$x=c$$ where $$c$$ is a constant has a gradient which is undefined.

Consider the gradient of the curve defined by $$y=f(x)$$ at the point P (ie the gradient of the tangent line AB).

This gradient cannot be calculated as only one point (the point $$P$$) on the line is known. But the point $$P$$ has coordinates $$\left(x,f(x)\right)$$and the point $$Q$$ has coordinates $$\left(x+h,f(x+h)\right)$$. The gradient of the line $$PQ$$ can be calculated and this can be used to approximate the gradient of $$AB$$. The gradient of $$PQ$$ = $$\frac{f(x+h)-f(x)}{h}$$. As the value of $$h$$ decreases (i.e $$Q$$ becomes closer to the point $$P$$), the approximation of the gradient is more accurate. The value of the gradient becomes most accurate as $$h$$ approaches zero.

The gradient formula for the curve $$y=f(x)$$ is defined as the derivative function

$f'(x)=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h},h\text{\neq }0.$

The derivative function$$f'(x)$$ gives the slope of the tangent to the curve$$f(x)$$ at any point x.

### Example

The derivative of the function $$f(x)=\frac{3}{x}$$ is $$f'(x)=-\frac{3}{x^{2}}$$. Find the slope of the tangent to the curve at $$x=4$$.

Solution

At $$x$$ = 4, \begin{align*} f'(x) & =-\frac{3}{4^{2}}\\ & =-\frac{3}{16}. \end{align*} Hence the slope of the tangent at $$x=4$$ is $$-\frac{3}{16}\,.$$

### Exercises

1. If the derivative function for $$f(x)=x^{3}-x$$ is $$f'(x)=3x^{2}-1$$, find the slope of the tangent to this curve at
$$\quad\text{a) }x=2\quad\text{b) x=0\quad\text{c) x=-9 .} }$$

#### Answer

$$\text{a) }11\quad\text{b) -1\quad\text{c) x=242\, .} }$$
1. If the derivative function of $$f(x)=\sin\left(x\right)$$ is $$f'(x)=\cos\left(x\right)$$ find the gradient of $$f(x)=\sin\left(x\right)$$ at
$$\quad\text{a) }x=0\quad\text{b) x=\pi/2\quad\text{c) x=3.5\, .} }$$

#### Answer

$$\text{a) }1\quad\text{b) 0\quad\text{c) -0.94 } }$$.
1. Determine $$\lim_{h\rightarrow0}\frac{(x+h)^{2}-x^{2}}{h}$$ and hence find the slope of the tangent to the curve $$y=x^{2}$$ at
$$\quad\text{a) }x=2\quad\text{b) x=0\quad\text{c) x=-9\, .} }$$

#### Answer

$$lim_{h\rightarrow0}\frac{(x+h)^{2}-x^{2}}{h}=2x$$.
$$\quad\text{a) }4\quad\text{b) x=0\quad\text{c) -18\, .} }$$

What's next... D3 Differentiation from first principles

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