Maths > Differentiation > 3.0 Properties of Differentiation
Differentiation
1.0 Differentiation
2.0 Some Basic Differentiation formulae
3.0 Properties of Differentiation
4.0 Derivative of Common Functions
4.1 Derivative of Trigonometric functions
4.2 Derivative of Inverse Trigonometric functions
4.3 Derivative of Exponential and Logarithmic Function
4.4 Derivative Of Hyperbolic function
4.5 Questions related to derivation of common functions
5.0 Explicit and Implicit form:
6.0 Parametric Differentiation
7.0 Differentiation of one function w.r.t other
8.0 Matrix Differentiation
9.0 Logarathimic Differentiation
10.0 Differentiation using substitution
3.2 Product Rule:
4.2 Derivative of Inverse Trigonometric functions
4.3 Derivative of Exponential and Logarithmic Function
4.4 Derivative Of Hyperbolic function
4.5 Questions related to derivation of common functions
$$ \frac{d}{dx} \left( f(x) g(x) \right) = g(x) \frac{d}{dx} f(x) + f(x) \frac{d}{dx} g(x) $$ Proof:
$$\begin{equation} \begin{aligned} {\left( {f\left( x \right)g\left( x \right)} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {f\left( {x + h} \right)g\left( {x + h} \right)} \right) - \left( {f\left( x \right)g\left( x \right)} \right)}}{h} \\ \end{aligned} \end{equation} $$ Add and subtract $f\left( {x + h} \right)g\left( x \right)$ in numerator,
$${\left( {f\left( x \right)g\left( x \right)} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {f\left( {x + h} \right)g\left( {x + h} \right)} \right) - f\left( {x + h} \right)g\left( x \right)\, + f\left( {x + h} \right)g\left( x \right)\, - \left( {f\left( x \right)g\left( x \right)} \right)}}{h} $$ $$ {\left( {f\left( x \right)g\left( x \right)} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {f\left( {x + h} \right)g\left( {x + h} \right) - f\left( {x + h} \right)g\left( x \right)} \right) + \left( {f\left( {x + h} \right)g\left( x \right)\, - f\left( x \right)g\left( x \right)} \right)}}{h} $$ $$ {\left( {f\left( x \right)g\left( x \right)} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right)\left( {g\left( {x + h} \right) - g\left( x \right)} \right) + g\left( x \right)\left( {f\left( {x + h} \right)\, - f\left( x \right)} \right)}}{h} $$ $$ {\left( {f\left( x \right)g\left( x \right)} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{f\left( {x + h} \right)\left( {g\left( {x + h} \right) - g\left( x \right)} \right)}}{h} + \frac{{g\left( x \right)\left( {f\left( {x + h} \right)\, - f\left( x \right)} \right)}}{h}} \right) $$ $$ {\left( {f\left( x \right)g\left( x \right)} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{f\left( {x + h} \right)\left( {g\left( {x + h} \right) - g\left( x \right)} \right)}}{h}} \right) + \mathop {\lim }\limits_{h \to 0} \left( {\frac{{g\left( x \right)\left( {f\left( {x + h} \right)\, - f\left( x \right)} \right)}}{h}} \right) $$ $$ {\left( {f\left( x \right)g\left( x \right)} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \left( {f\left( {x + h} \right)\frac{{\left( {g\left( {x + h} \right) - g\left( x \right)} \right)}}{h}} \right) + \mathop {\lim }\limits_{h \to 0} \left( {g\left( x \right)\frac{{\left( {f\left( {x + h} \right)\, - f\left( x \right)} \right)}}{h}} \right) $$ $$ {\left( {f\left( x \right)g\left( x \right)} \right)^\prime } = f\left( x \right)g'\left( x \right) + g\left( x \right)f'\left( x \right)$$
Example 4. $f(x)= x^3 $ and $g(x) = x^4$ prove $ \frac{d}{dx} \left( f(x) g(x) \right) = g(x) \frac{d}{dx} f(x) + f(x) \frac{d}{dx} g(x) $
Solution: Given: $$f(x)= x^3 $$ $$g(x) = x^4$$ $$f(x) g(x) = x^3 (x^4) = x^7$$
L.H.S: $$\frac{{d}}{{dx}}\left( f(x) g(x) \right) = \frac{d}{dx} x^7$$
Since we know that, $$\frac{d}{dx}x^n = nx^{n-1}$$
$$\frac{d}{dx} x^7 = 7x^{7-1} = 7x^6$$
R.H.S: $$g(x) \frac{d}{dx} f(x) + f(x) \frac{d}{dx} g(x) = x^4 \frac{d}{dx}x^3 + x^3 \frac{d}{dx}x^4 $$
Since we know that, $$\frac{d}{dx}x^n = nx^{n-1}$$
$$ x^4 \frac{d}{dx}x^3 + x^3 \frac{d}{dx}x^4 = x^4 (3x^{3-1}) + x^3 (4x^{4-1}) $$ $$x^4 \frac{d}{dx}x^3 + x^3 \frac{d}{dx}x^4 =x^4 (3x^2) + x^3(4x^3) = 3x^6+ 4x^6 =7x^6$$
thus, L.H.S = R.H.S
Note: $$ \frac{d}{dx} \left( f(x) g(x) \right) \ne \frac{d}{dx} f(x) \frac{d}{dx} g(x) $$
Example 5: $f(x)= 9x^3$ and $g(x) = 2$ prove $ \frac{d}{dx} \left( f(x) g(x) \right) \ne \frac{d}{dx} f(x) \frac{d}{dx} g(x) $
Solution: Given: $$f(x)= 9x^3 $$ $$g(x) = 2$$ $$f(x) g(x) = (9x^3)2 = 18 x^3$$
L.H.S: $$\frac{{d}}{{dx}}\left( f(x) g(x) \right) = \frac{d}{dx} 18x^3$$ Since, we know that $$\frac{d}{dx}x^n = nx^{n-1}$$
$$\frac{d}{dx} 18 x^3 = 18 (3)x^{3-1} = 54 x^2$$
R.H.S: $$ \frac{d}{dx} f(x) \frac{d}{dx} g(x) = \frac{d}{dx}(9x) \frac{d}{dx}(2) $$ Using Scalar diffeerentiation,
Also we know that, $$\frac{d}{dx}x^n = nx^{n-1}$$ $$\frac{d}{dx}k=0 \quad where \; k = \; constant $$ $$\frac{d}{dx}(9x) \frac{d}{dx}(2) = 9 \frac{d}{dx}x (0) $$ $$\frac{d}{dx}(9x) \frac{d}{dx}(2) = 9 x^{1-1} (0) = 9(0) = 0$$
Thus L.H.S $\ne$ R.H.S
