Rules of Differential Calculus
We already have understood how we calculate derivatives of some simple functions. However, in most of the questions that we will face in exam, things would probably not be so simple.
Many a times we will have to calculate derivatives of complex functions, which are made by combining simple functions in various ways.
For example, we know that:
$\frac{d}{dx} sin x = cos x$
$\frac{d}{dx} cos x = - sin x$
However, $\frac{d}{dx} sin x \hspace{1ex} cos x ≠ cos x (- sin x)$
We will have to follow some rules of Differential Calculus to find out derivatives of such complex expressions. Let’s see some of these rules.
Rules of Differential Calculus
We will represent functions f(x) and g(x) as f and g. And their derivatives f’(x) and g’(x) as f’ and g’.
Multiplication by constant Rule
$\frac{d}{dx} c f = c f'$
Q. What is $\frac{d}{dx} 5x^2$
Explanation:
$\frac{d}{dx} 5x^2$ = 5 $\frac{d}{dx} x^2$ = 5 (2x) = 10x
Sum and Difference Rule
$\frac{d}{dx} [f + g] = f’ + g'$
$\frac{d}{dx} [f - g] = f’ - g'$
Q. What is $\frac{d}{dx} (x^2 + sin x)$
Explanation:
$\frac{d}{dx} (x^2 + sin x)$ = $\frac{d}{dx} x^2 + \frac{d}{dx} sin x$ = 2x + cos x
Product and Division Rule
$\frac{d}{dx} [f . g] = f’ . g + f . g'$
$\frac{d}{dx} [f / g] = \frac{f’ . g - f . g’}{g^2}$
Special Case:
$\frac{d}{dx} [1 / f] = \frac{- f’}{f^2}$
Q. What is $\frac{d}{dx} x sin x$
Explanation:
$\frac{d}{dx} x sin x = x \frac{d}{dx} sin x + sin x \frac{d}{dx} x$ = x cos x + sin x
Composition of Functions Rule
$\frac{d}{dx} fºg = (f’ºg) . g'$
or $\frac{d}{dx} f(g) = f’(g) . g’
We can also represent the above in the following manner.
$\frac{dy}{dx} = \frac{dy}{du} . \frac{du}{dx}$
Q. What is $\frac{d}{dx} sin x^3$
Explanation:
Here, g = $x^3$, and so f(g) = sin g
So, f’(g) = cos g = cos $x^3$
g’ = $\frac{d}{dx} g = \frac{d}{dx} x^3 = 3x^2$
$\frac{d}{dx} sin x^3 = f’(g) . g’ = 3x^2 . cos x^3$