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Introductory Differentiation

Introductory Differentiation

Polynomials and more generally, terms of the form $ax^\nu$, where $a$ is a constant, can be differentiated by using the rule $$ \frac{d}{dx}x^\nu = \nu x^{\nu-1} $$ The operation of differentiation is linear, so $$ \frac{d}{dx}\left(af(x)\right) = a\frac{d}{dx}f(x)= af'(x) $$

Finally, we note the two base cases; $$\frac{d}{dx}(x) = 1,\; \frac{d}{dx}(a) = 0$$

Introductory Differentiation 1

Differentiate the polynomial $$P(x) = 4x^3 - 8x^2 + 6x + 2$$

solution - press button to display

$$\begin{align} P'(x) &= \frac{d}{dx}(P(x)) \\ &= \frac{d}{dx}(4x^3 - 8x^2 + 6x + 2) \\ &= \frac{d}{dx}(4x^3) - \frac{d}{dx}(8x^2) + \frac{d}{dx}(6x) + \frac{d}{dx}(2) \\ &= 12x^2 - 16x + 6 \end{align}$$

Introductory Differentiation 2

Differentiate the function, $f(x)$, given below $$ f(x) = \frac{x^2 + 2x - 3}{\sqrt{x}} $$
solution - press button to display

This expression must be manipulated before we can differentiate it using introductory differentiation techniques (we can do it directly using the quotient rule)

Let us rewrite $f(x)$ in the following manner

$$ \begin{align} f(x) &= \frac{x^2 + 2x - 3}{\sqrt{x}}\\ &= \frac{x^2 + 2x - 3}{x^\frac{1}{2}} \\ &= x^{\frac{3}{2}} + 2x^\frac{1}{2} - 3x^{-\frac{1}{2}} \end{align} $$

 

This function is now in the form where we can directly differentiate it.

$$ \begin{align} f'(x) &= \frac{d}{dx}\left(x^{\frac{3}{2}} + 2x^\frac{1}{2} - 3x^{-\frac{1}{2}}\right)\\ &= \frac{d}{dx}\left(x^{\frac{3}{2}}\right) + \frac{d}{dx}\left(2x^\frac{1}{2}\right) - \frac{d}{dx}\left(3x^{-\frac{1}{2}}\right)\\ &= \frac{3}{2}x^\frac{1}{2} + x^{-\frac{1}{2}} + \frac{3}{2}x^{-\frac{3}{2}} \end{align} $$