In calculus (a branch of mathematics) the derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one
quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a vehicle with respect to time is the instantaneous velocity at which the
vehicle is traveling.

Before solving them we need to know some formula’s which will help us to evaluate the problems.

The formulae are:

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

(2) `d/dx (1/x^n) = -n / x^[n +1]` .

(3) `d/dx ` (e^{x}) = e^{x}.

(4) `d/dx` (`sqrtx` ) = `1/[2 sqrtx]` .

(5) `d/dx` (log x) = `1/x` .

(6)` d/dx` (sinx) = cosx.

(7)` d/dx ` (cosx) = -sinx.

(8) `d/dx` (tanx) = sec^{2}x.

(9) `d/dx` (secx) = secx tanx.

(10) `d/dx` (cotx) = -cosec^{2}x.

(11) `d/dx` (cosecx) = - cosecx cotx.

Before we attempt to solve any of the problems on this topic, let us remember all the above formulas, so that the calculation will go very smoothly.

**Differentiation by parts problem 1:**

Find the value of (dy / dx). y = (19x + 3) (4x^{2} + 31x)

**Solution:**

Given function y = (19x + 3) (4x^{2} + 31x)

The above has been separated by two parts, they are

f(x) = (19x + 3) and g(x) = (4x^{2} +
31x)

**Formula:**

(d / dx) [f(x) g(x)] = f(x) g'(x) + g(x) f'(x)

Differentiate the given function with respect to x, we get

f'(x) = 19 and g'(x) = 8x + 31

Substitute the values in the above formula, we get

(dy / dx) = (19x + 3) (8x + 31) + (4x^{2} + 31x) (19)

= 152x^{2} + 589x +
24x + 93 + 76x^{2} + 589x

= 228x^{2} + 1202x
+ 93

**Answer:**

The final answer is 228x^{2} + 1202x + 93

**Differentiation by parts problem 2:**

Find the value of (dy / dx). y = (x cosx)

**Solution:**

Given function y = (x cosx)

The above has been separated by two parts, they are

f(x) = x and g(x) = cosx

**Formula:**

(d / dx) [f(x) g(x)] = f(x) g'(x) + g(x) f'(x)

Differentiate the given function with respect to x, we get

f'(x) = 1and g'(x) = - sinx

Substitute the values in the above formula, we get

(dy / dx) = x (- sinx) + (cosx) (1)

= - x sinx + cosx

**Answer:**

The final answer is - x sinx + cosx

**Differentiation by parts problem 3:**

** **

Find the value of (dy / dx). y = (x^{2}) * (3sinx)

**Solution:**

Given function y = (x^{2}) * (3sinx)

The above has been separated by two parts, they are

f(x)
= (x^{2}) and g(x) = (3sinx)

**Formula:**

(d / dx) [f(x) g(x)] = f(x) g'(x) + g(x) f'(x)

Differentiate the given function with respect to x, we get

f'(x) = 2x and g'(x) = 3cosx

Substitute the values in the above formula, we get

(dy / dx) = x^{2} (3cosx) + (3sinx) (2x)

= 3x^{2}cosx + 6x
sinx

**Answer:**

The final answer is 3x^{2}cosx + 6x sinx

**Differentiation by parts problem 1:**

** **

** **Differentiate the given function f(x) = 4x (9x^{2} + 3x)

**Answer:**

The final answer is 108x^{2} + 24x