# What Is The Derivative Of F(X)= 5 Derivatives, Derivative Definition

**What Is The Derivative Of F(X)= 5 Derivatives, Derivative Definition**in

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The good news is we can find the derivatives of polynomial expressions without using the delta method that we met in The Derivative from First Principles.

Isaac Newton and Gottfried Leibniz obtained these rules in the early 18th century. They follow from the “first principles” approach to differentiating, and make life much easier for us.

## Common derivatives

### a. Derivative of a Constant

`(dc)/(dx)=0`

This is basic. In English, it means that if a quantity has a constant value, then the rate of change is zero.

Example a

`d/dx(6)=0`

### b. Derivative of n-th power of x

`d/(dx)x^n=nx^(n-1)`

This follows from the delta method.

Example b

`d/(dx)x^5=5x^4`

### c. Derivative of Constant product

`d/(dx)(cy)=c(d)/(dx)(y)=c(dy)/dx`

Here, *y* is some function of *x*. It means that if we are finding the derivative of a constant times that function, it is the same as finding the derivative of the function first, then multiplying by the constant.

Example c

If `y=x^7`, then `dy/dx=d/dx(x^7)=7x^6`.

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Applying the new rule (c), we have:

`d/(dx)(3y)=d/(dx)(3x^7)`

`=3(d)/(dx)(x^7)`

`=3(dy)/dx`

`=(3)(7x^6)`

`=21x^6`

### d. Derivative of a sum

`(d(u+v))/(dx)=(du)/(dx)+(dv)/(dx)`

Here, *u* and *v* are functions of *x*. The derivative of the sum is simply equal to the derivative of the first plus derivative of the second. It does not work the same for the derivative of the product of two functions, that we meet in the next section.

Example d

If `u=x^2` and `v=x^9`, then:

`(d(u+v))/(dx)=d/dx(x^2)+d/dx(x^9)`

`=2x+9x^8`

## Derivatives Summary

Constant: | `(dc)/(dx)=0` |

n-th power of x: | `d/(dx)x^n=nx^(n-1` |

Constant product: | `d/(dx)(cy)` `=c(d)/(dx)(y)=c(dy)/dx` |

Sum: | `(d(u+v))/(dx)=(du)/(dx)+(dv)/(dx)` |

## Further Examples

### Example 1

Find the derivative of y = −7×6

Answer

Using the rule

`d/(dx)(cy)=c(d)/(dx)(y)`

we can take the -7 out the front:

`d/(dx)(-7x^6)=-7d/(dx)(x^6)`

And

`d/(dx)x^n=nx^(n-1)`

gives us:

`-7d/(dx)(x^6)=-7xx6x^5=-42x^5`

Note: We can do this in one step:

`(dy)/(dx)=-42x^5`

We can write: `(dy)/(dx)=-42x^5` OR `y”=-42x^5`. They mean the same thing.

### Example 2

Find the derivative of y = 3×5 − 1

Answer

`y = 3x^5− 1`

Now,

`d/(dx)(3x^5)=3xx5x^4=15x^4`

And since `(dc)/(dx)=0`, we can write:

`d/(dx)(-1)=0`

So

`(dy)/(dx)=d/(dx)(3x^5-1)=15x^4`

### Example 3

Find the derivative of

`y=13x^4-6x^3-x-1`

Answer

Now, taking each term in turn:

`d/(dx)(13x^4)=52x^3` (using `d/(dx)x^n=nx^(n-1)`)

`d/(dx)(-6x^3)=-18x^2` (using `d/(dx)x^n=nx^(n-1)`)

`d/(dx)(-x)=-1` (since `-x = -(x^1)` and so the derivative will be `-(x^0) = -1`)

`d/(dx)(-1)=0` (since `(dc)/(dx)=0`)

So

`(dy)/(dx)=52x^3-18x^2-1`

### Example 4

Find the derivative of

`y=-1/4x^8+1/2x^4-3^2`

Answer

`y=-1/4x^8+1/2x^4-3^2`

Differentiating term by term, we have:

`d/(dx)(-1/4x^8)=-8/4x^7=-2x^7`

`d/(dx)(1/2x^4)=4/2x^3=2x^3`

`d/(dx)(3^2)=0` (this is the derivative of a constant)

So

`(dy)/(dx)=d/(dx)(-1/4x^8+1/2x^4-3^2)` `=-2x^7+2x^3`

### Example 5

Evaluate the derivative of

`y=x^4-9x^2-5x`

at the point `(3,-15)`.

Answer

`y=x^4-9x^2-5x`

So

`(dy)/(dx)=4x^3-18x-5`

At the point where `x = 3`, the derivative has value:

`(dy)/(dx)=4(3)^3-18(3)-5`

`=4xx27-18xx3-5`

`=49`

This means that the slope of the curve `y=x^4-9x^2-5x` at* *`x= 3` is `49`.

Here is a graph of the curve showing the slope we just found. (The axes are not scaled the same.)

The curve `y=x^4-9x^2-5x` showing the tangent at `(3,-15).`

### Example 6

Find the derivative of the function

`y=x^(1//4)-2/x`

Answer

In this case we have fractions and negative numbers for the powers of x. (So it is not a polynomial).

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The differentiation rules still apply.

`y=x^(1//4)-2/x`

We can write this as:

`y=x^(1//4)-2x^-1`

Differentiating gives us:

`(dy)/(dx)=1/4x^(1/4-1)-2(-1)x^(-1-1`

`=1/4x^(-3//4)+2x^-2`

`=1/(4x^(3//4))+2/x^2`

## Exercise

Find the equation of the tangent to the curve `y = 3x − x^3` at `x = 2`.

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Answer

Now `y = 3x − x^3`

`(dy)/(dx)=3-3x^2` and the value of this derivative at `x=2` is given by:

`(dy)/(dx)=3-3(2)^2=-9`

Since `y = 3x − x^3`,* *then when `x= 2`, `y= -2.`

So we need the equation of the line passing through `(2,-2)` with slope `-9`.

Using the general equation of the line `y-y_1=m(x-x_1)`, we have:

`y + 2 = -9(x – 2)`

So the required equation is:

`y = -9x + 16`

Or, in general form: `9x + y – 16 = 0.`

Here”s the graph of the situation:

The curve `y = 3x − x^3` showing the tangent at `(2, -2)`

**Derivative**