2. Application of the First Principles

L2MAEL

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Application of the Differentiation Formula:

Using that exact formula, $f(x)=x^2$ will be differentiated below:

$$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{(x+h{)}^2-x^2}{h}$$

The $x$ in the formula, is replaced with the function, which in this case is $x^2$.

$$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{x^2+2xh+h^2-x^2}{h}$$ $$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{2xh+h^2}{h}$$ $$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{h(2x+h)}{h}$$ $$f^{\prime }(x)=\underset{h\to 0}{\lim }2x+h$$

and since $h$ is so close to zero, we can basically ‘treat’ $h$ as zero here:

$$f^{\prime }(x)=2x$$

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Differentiating $f(x)=x^3$:

$$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{(x+h{)}^3-x^3}{h}$$ $$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{x^3+3h{x}^2+3x{h}^2+h^3}{h}$$ $$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{x^3+3h{x}^2+3x{h}^2+h^3-x^3}{h}$$ $$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{3h{x}^2+3x{h}^2+h^3}{h}$$ $$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{h(3x^2+3xh+h^2)}{h}$$ $$f^{\prime }(x)=\underset{h\to 0}{\lim }3x^2+3xh+h^2$$

treating $h$ as zero:

$$f^{\prime }(x)=3x^2$$

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Differentiating $f(x)=3x^2+2x-1$:

$$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{3(x+h{)}^2+2(x+h)-1-(3x^2+2x-1)}{h}$$ $$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{3(x^2+2hx+h^2)+2x+2h-1-3x^2-2x+1}{h}$$ $$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{6xh+3h^2+2h}{h}$$ $$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{h(6x+3h+2)}{h}$$ $$f^{\prime }(x)=\underset{h\to 0}{\lim }6x+3h+2$$

treating $h$ as zero:

$$f^{\prime }(x)=6x+2$$

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Differentiating $f(x)=\frac{5}{x}$:

$$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{\frac{5}{(x+h)}-\frac{5}{x}}{h}$$ $$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{\frac{5x-5x-5h}{x^2+hx}}{\frac{h}{1}}$$ $$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{\frac{-5h}{x^2+hx}}{\frac{h}{1}}$$

fraction over fraction calculation:

$$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{-5h}{h{x}^2+h^{2}x}$$ $$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{-5h}{h(x^2+hx)}$$ $$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{-5}{(x^2+hx)}$$

treating $h$ as zero:

$$f^{\prime }(x)=\frac{-5}{x^2}$$

Comparing: $f(x)$ was $5x^{-1}$ $f^{\prime }(x)$ is $-5x^{-2}$

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Pattern in the Formula:

Using the formula above for long equations is difficult, for example:

$$f(x)=5x^4+3x-5+x^{-3}+4\sqrt[3]{x}+\frac{5}{x^2}-\frac{1}{\sqrt[4]{x^3}}$$

Therefore, using the pattern defined below to differentiate is a much faster way, and would reduce mistakes.

The exponent comes down, and multiplies with the coefficient (in this case, it’s 1, so nothing happens), then the exponent decreases by one. Since $2\times 1=2$, and $2-1=1$, the coefficient turns into 2, and the exponent turns into 1. Therefore, the differentiated form would be: $2x$

This same principle applies with all other functions.

The formula for differentiating with regards to the pattern would be:

$$\frac{d}{dx}(a{x}^n)=n\times a\times x^{n-1}$$