## unimatix one year ago Derivatives without using shortcut method. Problem: g(x) = x sqrt(x)

1. unimatix

Going to show what I have as of now: $g(x) = xsqrt(x)$ $y = xsqrt(x)$ $y + \triangle y = x + \triangle(x)\sqrt{ (x+\triangle x ) )}$

2. unimatix

|dw:1441578438052:dw|

3. triciaal

|dw:1441579069139:dw|

4. triciaal

|dw:1441579234601:dw|

5. triciaal

this is going too far back

6. unimatix

I know the product rule. I'm trying to do it without though.

7. unimatix

I tried to rationalize the numerator on the left and got nothing.

8. anonymous

$g(x) = x\sqrt{x}$$g(x+h) = (x+h)(\sqrt{x+h})$$\frac{g(x+h)-g(x)}{h}=\frac{(x+h)(\sqrt{x+h})-x\sqrt{x}}{h}$

9. anonymous

Meh.. the only thing that'll work here would be : $=\frac{\sqrt{(x+h)^3}-x^{3/2}}{h}= \frac{(x+h)^{3/2}-x^{3/2}}{h}$

10. amistre64

you are trying to work first principles .. the limit definition of a derivative right?

11. anonymous

Mmhmm... $$\lim_{h\rightarrow 0}$$

12. amistre64

(x+h) sqrt(x+h) - x sqrt(x) ------------------------- h x sqrt(x+h) +hsqrt(x+h) - x sqrt(x) ------------------------------- h x (sqrt(x+h) - sqrt(x)) sqrt(x+h) + ------------------- h x (h) sqrt(x+h) + ------------------- h(sqrt(x+h) + sqrt(x)) x sqrt(x+h) + ------------------- sqrt(x+h) + sqrt(x))

13. amistre64

14. anonymous

Not the third step.

15. amistre64

(ac+b)/a = c + (b/a)

16. amistre64

you just split the fraction into convenient terms, and the left side is just a simplified result of h/h sqrt(x+h)

17. anonymous

Oh, I see what you did now.

18. amistre64

$\frac{a+bh+c}{h}\color{red}{\implies}\frac{bh}{h}+\frac{a+c}{h}$

19. amistre64

the rest is taken care of by a conjugate

20. anonymous

Just trying to solve it by myself real quick and cross-checking between my work and your method :)

21. anonymous

Then.... $\lim_{h\rightarrow 0} \left(\sqrt{x+h} +\frac{x}{2\sqrt{x}}\right)$$=\sqrt{x}+\frac{x}{2\sqrt{x}}$$=\frac{2x+x}{2\sqrt{x}}$$=\frac{3x}{2\sqrt{x}}$$=\frac{3x^{3/2}}{2x}$$=\boxed{\frac{3}{2}x^{1/2}}$