Power Rule of Derivatives – Examples and Practice Problems

We start by writing the formula for the power rule:

$latex f'(x^n) = nx^{n-1}$

In this case, we have the exponent $latex n=4$. Then, by using the power rule, we have:

$$\frac{d}{dx} (x^n) = \frac{d}{dx} (x^4)$$

$$\frac{d}{dx} (x^4) = 4 \cdot x^{4-1}$$

Simplifying this, we have:

$$\frac{d}{dx} (x^4) = 4 x^3$$

$$f'(x)= 4 x^3$$

Let’s solve using the formula for the power rule of derivatives:

$latex f'(x^n) = nx^{n-1}$

We can identify the exponent $latex n=3$. Then, by applying the power rule, we have:

$$\frac{d}{dx} (x^n) = \frac{d}{dx} (x^3)$$

$$\frac{d}{dx} (4x^3) = 3 \cdot (4x^{3-1})$$

When we simplify the expression, we have:

$$\frac{d}{dx} (4x^3) = 12 x^2$$

$$f'(x)= 12 x^2$$

The formula for the power rule of derivatives is:

$latex f'(x^n) = nx^{n-1}$

In this case, we have the exponent $latex n=8$. Applying the power rule, we have:

$$\frac{d}{dx} (x^n) = \frac{d}{dx} (7x^8)$$

$$\frac{d}{dx} (7x^8) = 8 \cdot (7x^{8-1})$$

When we simplify this expression, we have:

$$\frac{d}{dx} (7x^8) = 56 x^7$$

$$f'(x)= 56 x^7$$

We start by writing the formula for the power rule:

$latex f'(x^n) = nx^{n-1}$

In this case, we have a negative exponent $latex n=-5$. However, we simply have to apply the power rule as in the previous cases:

$$\frac{d}{dx} (x^n) = \frac{d}{dx} (3x^{-5})$$

$$\frac{d}{dx} (3x^{-5}) = -5 \cdot (3x^{-5-1})$$

When we simplify the expression, we have:

$$\frac{d}{dx} (3x^{-5}) = -15 x^{-6}$$

$$f'(x)= -15 x^{-6}$$

We can use the laws of exponents to write as follows:

$$f'(x) = -\frac{15}{x^6}$$

Let’s use the power rule of derivatives:

$latex f'(x^n) = nx^{n-1}$

We see that the exponent is $latex n=-6$. Therefore, using the power rule with this function, we have:

$$\frac{d}{dx} (x^n) = \frac{d}{dx} (-5x^{-6})$$

$$\frac{d}{dx} (-5x^{-6}) = -6 \cdot (-5x^{-6-1})$$

When we simplify, we have:

$$\frac{d}{dx} (-5x^{-6}) = 30 x^{-7}$$

$$f'(x)= 30 x^{-7}$$

We can write as follows for simplicity:

$$f'(x) = \frac{30}{x^7}$$

The formula for the power rule is:

$latex f'(x^n) = nx^{n-1}$

In this case, we have a variable in the denominator. Therefore, we can use the rules of exponents to write as follows:

$latex f(x)=x^{-5}$

Then, we see that we have the exponent $latex n=-5$. Using the power rule, we have:

$$\frac{d}{dx} (x^n) = \frac{d}{dx} (x^{-5})$$

$$\frac{d}{dx} (x^{-5}) = -5 \cdot x^{-5-1}$$

Simplifying this, we have:

$$\frac{d}{dx} (x^{-5}) = -5 x^{-6}$$

$$f'(x)= -5 x^{-6}$$

We use the rules of exponents again to write as follows:

$$f'(x) = -\frac{5}{x^6}$$

We start by writing the formula for the power rule:

$latex f'(x^n) = nx^{n-1}$

We use the rules of exponents to write like this:

$latex f(x)=\frac{2}{3} x^{-2}$

Now, we have the exponent $latex n=-2$. Using the power rule, we have:

$$\frac{d}{dx} (x^n) = \frac{d}{dx} (\frac{2}{3} x^{-2})$$

$$\frac{d}{dx} (\frac{2}{3} x^{-2}) = -2 \cdot (\frac{2}{3} x^{-2-1})$$

Simplifying this, we have:

$$\frac{d}{dx} (\frac{2}{3} x^{-2}) = -\frac{4}{3} x^{-3}$$

$$f'(x)= -\frac{4}{3} x^{-3}$$

Finally, we can write as follows:

$$f'(x)= -\frac{4}{3x^3}$$

The power rule of derivatives is:

$latex f'(x^n) = nx^{n-1}$

In this case, we have the exponent $latex n=\frac{1}{3}$. When we use the power rule, we have:

$$\frac{d}{dx} (x^n) = \frac{d}{dx} (x^{\frac{1}{3}})$$

$$\frac{d}{dx} (x^{\frac{1}{3}}) = \frac{1}{3} \cdot (x^{\frac{1}{3}-1})$$

Simplifying this, we have:

$$\frac{d}{dx} (x^{\frac{1}{3}}) = \frac{1}{3} x^{-\frac{2}{3}}$$

$$f'(x)= \frac{1}{3} x^{-\frac{2}{3}}$$

Using the rules of exponents, we can write as follows:

$$f'(x)= \frac{1}{3 x^{\frac{2}{3}}}$$

$$f'(x)= \frac{1}{3 \sqrt[3]{x^2}}$$

We have the following rule:

$latex f'(x^n) = nx^{n-1}$

We can use the rules of exponents to rewrite the radical:

$$ \sqrt{x}=x^{\frac{1}{2}}$$

Now, we see that the exponent is $latex n = \frac{1}{2}$. Then, using the power rule on the function, we have:

$$\frac{d}{dx} (x^n) = \frac{d}{dx} (x^{\frac{1}{2}})$$

$$\frac{d}{dx} (x^{\frac{1}{2}}) = \frac{1}{2} \cdot ( x^{\frac{1}{2}-1})$$

Simplifying, we have:

$$f'(x) = \frac{1}{2} x^{-\frac{1}{2}}$$

Using the rules of exponents, we write as follows:

$$f'(x) = \frac{1}{2x^{\frac{1}{2}}}$$

$$f'(x) = \frac{1}{2\sqrt{x}}$$

We have the following rule:

$latex f'(x^n) = nx^{n-1}$

We use the rules of exponents to write the function like this:

$$ \frac{1}{\sqrt{x}}=x^{-\frac{1}{2}}$$

The exponent of the function is $latex n = -\frac{1}{2}$. Therefore, by applying the power rule of derivatives on the function, we have:

$$\frac{d}{dx} (x^n) = \frac{d}{dx} (x^{-\frac{1}{2}})$$

$$\frac{d}{dx} (x^{-\frac{1}{2}}) = -\frac{1}{2} \cdot ( x^{-\frac{1}{2}-1})$$

Simplifying, we have:

$$f'(x) = -\frac{1}{2} x^{-\frac{3}{2}}$$

We can use the rules of exponents again to write:

$$f'(x) = -\frac{1}{2x^{\frac{3}{2}}}$$

$$f'(x) = -\frac{1}{2\sqrt{x^3}}$$