# SUCCESSIVE DIFFERENTIATION

Successive differentiation is the process of differentiating a given function successively. Through successive differentiation, we can obtain the first, second, third, and other derivatives of a function.

Thus, the function $y=f(x)$ can be differentiated to x, and the differential coefficient is written as $\frac{dy}{dx}$ or $f'(x)$.

If the function is differentiated again, the second differential coefficient is obtained and it is written as $\frac{d^2y}{dx^2}$ or $f''(x)$.

By successive differentiation, Further derivatives such as $\frac{d^3y}{dx^3}$,$\frac{d^4y}{dx^4}$ and $\frac{d^5y}{dx^5}$ can be obtained

For example, If $y=12x^3$, Then

$$\frac{dy}{dx}=36x^2$$

$$\frac{d^2y}{dx^2}=72x$$

$$\frac{d^3y}{dx^3}=72$$

$$\frac{d^4y}{dx^4}=0$$

Example 1

if $y=x^4+6x^3-5x^{-2}$, Evaluate $\frac{d^2y}{dx^2}$

Solution:

$\frac{d^2y}{dx^2}$ means we should solve for the second derivative of the function.

$$\frac{dy}{dx}=4x^3+18x^2+10x^{-3}$$

Now, let's take the second derivative

$$\frac{d^2y}{dx^2}=12x^2+36x-\frac{30}{x^4}$$

Example 2

If $y=\sin (3x)$, Find the third derivative of the function.

Solution:

Recalled, to differentiate composite trigonometric functions, we use;

$f(g(x))=f'(g(x))\times g(x)$

Here $f'=\cos$, $g(x)=3x$, $g'(x)=3$, Therefore,

$$\frac{dy}{dx}=\cos (3x)\times 3$$

$$\frac{dy}{dx}=3\cos (3x)$$

Now, let take the second derivative using the same logic.

$$\frac{d^2y}{dx^2}=-3\sin (3x)\times 3$$

$$\frac{d^2y}{dx^2}=-9\sin (3x)$$

Now, to the third derivative

$$\frac{d^3y}{dx^3}=-9\cos (3x)\times 3$$

$$\frac{d^3y}{dx^3}=-27\cos (3x)$$

Example 3

if $h(x)=e^x$, evaluate $h''''(x)$

Solution:

$h''''(x)$ means we should find the fourth derivative.

let find the first derivative.

Recall, To differentiate exponential functions, We use:

$$a^{f(x)}=a^{f(x)}\times \ln(a)\times of(x)$$

Here, $a=e$, $f(x)=x$ and $f'(x)=1S $$h'(x)=e^x \times \ln(e) \times 1$$ Recalled that$\ln (e)=1$$$h'(x)=e^x Now, let take the second derivative by the same reasoning is$$h''(x)=e^x \times \ln(e) \times 1h''(x)=e^x$$Taking the third derivative$$h'''(x)=e^x \times \ln(e) \times 1h'''(x)=e^x$$To the fourth derivative$$h''''(x)=e^x \times \ln(e) \times 1h''''(x)=e^x$$Example 4 Find \frac{d^4y}{dx^4} of y=\ln(x)+\ln(x^2) Solution: Using the property of logs, y=\ln(x)+\ln(x^2) can be rewritten as:$$y=\ln(x)+2 \ln(x)y=3 \ln(x)$$As you may recall, to differentiate natural logarithms, it's$$\frac{d}{dx}(\ln (f(x))=\frac{f'(x)}{f(x)}\frac{dy}{dx}=\frac{3}{x}=3x^{-1}$$Now, we differentiate again using the power rule$$\frac{d^2y}{dx^2}=-3x^{-2}$$Again, we differentiate$$\frac{d^3y}{dx^3}=6x^{-3}$$Differentiating for the fourth time.$$\frac{d^4y}{dx^4}=-18x^{-4}\frac{d^4y}{dx^4}=\frac{-18}{x^4}$\$

There you have it. Got questions relating to this post? Feel free to ask our telegram community.

Meanwhile, if you would like to know more about differentiation, Check out the following.

Help us grow our readership by sharing this post