can someone show me how to differentiate stuff like x + 2 and
I've never did this before and I use the most god awful textbook imaginable.
Much thanks
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$\begingroup$By definition the derivative $f'$ of a function $f: \mathbb{R} \to \mathbb{R}$ is $$f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}.$$ Fix any real $h > 0$. Since the quotient $$\dfrac{(x+h+2) - (x+2)}{h} = 1,$$ we have immediately $$\lim_{h \to 0} \dfrac{(x+h+2) - (x+2)}{h} = 1,$$ which is exactly the derivative of the function $x \mapsto x+2$.
$\endgroup$ 8 $\begingroup$- In general the derivative of a sum is the sum of the derivatives.
- The derivative of something of the form $ax^b$ is $abx^{b-1}$
Examples: $$\frac{d(c)}{dx}=\frac{d(cx^0)}{dx}=c \cdot 0 x^{0-1}=0$$ Where $c$ is a constant. $$\frac{d(x^2+3x+4)}{dx}=\frac{d(x^2)}{dx}+\frac{d(3x)}{dx}+\frac{d(4)}{dx}=2x^{2-1}+3\cdot 1 \cdot x^{1-1}+0=2x+3$$
I think you can continue with your exercises at that point
$\endgroup$ $\begingroup$Hint:$$ \frac{d}{dx}[f(x)\pm g(x)]=\frac{d}{dx}[f(x)]\pm \frac{d}{dx}[g(x)] $$ $$ \frac{d}{dx}[ax^n]= a\frac{d}{dx}[x^n]= anx^{n-1} $$
$\endgroup$ 2 $\begingroup$Two things to know:
- the derivative of a linear combination is the linear combination of the derivatives,
$$(x^6-5x^5-2x^4)'=(x^6)'-5(x^5)'-2(x^4)'$$
- the derivative of a power is a power with exponent minus one, times the exponent.
$$(x^6)'=6x^5,$$ $$(x^5)'=5x^4,$$ $$(x^4)'=4x^3.$$
Why is it so ?
By definition, $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h.$$ For brevity, we will just write $$\frac{f_h-f_0}h,$$ where the subscript $_h$ means "evaluated at $x+h$" and the subscript $_0$, "evaluated at $x$".
$$\frac{(af+bg)_h-(af+bg)_0}h=\frac{(af_h+bg_h)-(af_0+bg_0)}h=a\frac{f_h-f_0}h+b\frac{g_h-g_0}h,$$hence $(af+bg)'=af'+bg'$.
$$\frac{x^n_h-x^n_0}h=\frac{(x_h-x_0)(x^{n-1}_h+x^{n-2}_hx_0+x^{n-3}_hx^2_0+...x^{n-1}_0)}h=\\ x^{n-1}_h+x^{n-2}_hx_0+x^{n-3}_hx^2_0+...x^{n-1}_0.$$ This is obtained by factoring the binomial (you can check by executing the product and simplifying).
When $h$ tends to $0$, $x_h=x_0=x$, all $n$ terms become $x^{n-1}$, hence $(x^n)'=nx^{n-1}$.
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