Given a function *y* = ƒ( *x*), its **derivative**, denoted by *y*′ or *dy*/ *dx*, gives the instantaneous rate of change of ƒ with repect to *x*. Geometrically, this gives the slope of the curve (that is, the slope of the tangent line to the curve) *y* = ƒ( *x*).

Figure 1

The second derivative identifies the **concavity** of the curve *y* = ƒ( *x*). A portion of a differentiable curve *y* = ƒ( *x*) from *x* = *a* to *x* = *b* is said to be **concave up** if the curve lies above its tangent lines between *a* and *b*, and **concave down** if it lies below its tangent lines.

A curve *y* = ƒ ( *x*) is concave up at those points *x* where the second derivative is positive, and concave down where the second derivative is negative. Points where the concavity changes are called **inflection points** and are located at those points *x* _{0} where ƒ ^{n }( *x* _{0}) = 0 but ƒ ^{m }( *x* _{0}) ≠ 0.

Figure 2

Table 1

**TABLE 1 Computational Properties of Differentiation and Integration**

In addition to being familiar with the definitions and fundamental properties, you should, of course, be able to actually differentiate a function. Although Table *c* has been omitted from each of the integration formulas, as in Table 1

**TABLE 2**

**Differentiation and Integration Formulas**

**Example 1**: Differentiate each of the following:

a

. y= 3x^{2}− 5x+ 8b

. y=x^{2}e^{x }c

. y= Inx/xd

. y= (x^{3}+x− 1)^{4}e.

f

. y= sin(x^{2})g

. y= sin^{2}xh.

y= e^{tan x }i.

The solutions are as follows:

a

. y′ = 6x− 5b. Using the product rule,

y′ =x^{2}· e^{x }·e^{x }· 2x=xe^{x }(x+ 2)c. By the quotient rule,

All of the remaining parts use the chain rule (as embodied in the formulas in

). d

. y′ = 4(x^{3}+x− 1)^{3}· (3x^{2}+ 1)e.

f

. y′ = 2xcos(x^{2})g

. y= (sinx)^{2}⟹y′ = 2 sinxcosx= sin 2xh

. y′ = e^{tan x }sec^{2}xi.

**Example 2**: What is the equation of the tangent line to the curve *y* = *e* ^{x }In *x* at the point (1, 0)?

The first step is to find the slope of the tangent line at *x* = 1, which is the value of the derivative of *y* at this point:

slope at point

Since the **point‐slope formula** says that the straight line with slope **m** which passes through the point ( *x* _{0}, *y* _{0}) has the equation

the equation of the desired tangent line is *y* = *e*( *x* −1).

**Example 3:** Is the curve *y* = arcsin concave up or is it concave down at the point ( ?

Concavity is determined by the sign of the second derivative.

Since

the first derivative of *y* = arcsin is

Its second derivative is therefore

For *x* + ¼, the denominator in the expression above for *y*″ is positive (as it is for any *x* in the interval 0 < *x* < 1), but the numerator is negative. Therefore, *y*″(¼) < 0), and the curve is concave down at the point .

**Example 4:** consider the curve given implicitly by the equation

What is the slope of this curve at the point where it crosses the *x* axis?

To find the slope of a curve defined implicitly (as is the case here), the technique of **implicit differentiation** is used: Differentiate both sides of the equation with respect to *x*; then solve the resulting equation for *y*′.

The curve crosses the *x* axis when *y* = 0, and the given equation clearly implies that *x* = − 1 at *y* = 0. From the expression directly above, the slope of the curve at the point (−1, 0) is