## Calculus: Early Transcendentals (2nd Edition)

$a.$ $(x^2)'=2x$ $b.$ $(x^3)'=3x^2$ $c.$ $(x^4)'=4x^3$ $d.$ If $f(x)=x^n$ it looks like $f'(x)=nx^{n-1}$
We will use the given definition of the derivative in the problem. $a.$ $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{(x+h)^2-x^2}{h}=\lim_{h\to0}\frac{x^2+h^2+2xh-x^2}{h}=\lim_{h\to0}\frac{h^2+2xh}{h}=\lim_{h\to0}(h+2x)=0+2x=2x.$$ $b.$ $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{(x+h)^3-x^3}{h}=\lim_{h\to0}\frac{x^3+3x^2h+3xh^2+h^3-x^3}{h}=\lim_{h\to0}\frac{3x^2h+3xh^2+h^3}{h}=\lim_{h\to0}(3x^2+3xh+h^2)=3x^2+3x\cdot0+0^2=3x^2.$$ $c.$ $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{(x+h)^4-x^4}{h}.$$ We will use the transformation $$(x+h)^4-x^4=((x+h)^2+x^2)((x+h)^2-x^2)=((x+h)^2+x^2)(x+h+x)(x+h-x)=((x+h)^2+x^2)(2x+h)h.$$ Putting this into the limit we get $$f'(x)=\lim_{h\to0}\frac{((x+h)^2+x^2)(2x+h)h}{h}=\lim_{h\to0}((x+h)^2+x^2)(2x+h)=((x+0)^2+x^2)(2x+0)=4x^3.$$ $d.$ The pattern is $(x^2)'=2x,$ $(x^3)'=3x^2,$ $(x^4)'=4x^3$. It looks like when $f(x)=x^n$ then $f'(x)=nx^{n-1}.$