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

$$h''(x)=e^x \times \ln(e) \times 1$$

$$h''(x)=e^x$$

Taking the third derivative

$$h'''(x)=e^x \times \ln(e) \times 1$$

$$h'''(x)=e^x$$

To the fourth derivative

$$h''''(x)=e^x \times \ln(e) \times 1$$

$$h''''(x)=e^x$$

__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.

$$\frac{d^4y}{dx^4}=-18x^{-4}$$

$$\frac{d^4y}{dx^4}=\frac{-18}{x^4}$$

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