Derivative/ Trig Help!!!!

[DancingQueen09]

I need help on two questions! I have been trying them over and over again and still I cannot complete them. Thanks to everyone who helps.

1. Y= (x)(x^(1/2)) + (1)/((x^2)(x^1/2))

Find y'

The question is x radical x plus one over x^2 radical x

I know you use product rule and quotient rule then add them together but I still cant figure this out!!!
If you could please show step by step this will greatly help


2. Y= xSinx + (Sinx)/(1+Cosx)

Find A) f ' (x)

B) f '' (pi/2)


I also know you use product rule and quotient rule then add them together but I still cant figure this out!!!
If you could please show step by step this will greatly help

 

[galactus]

I am sorry, I think I misread your problem.

Is it $\displaystyle x\sqrt{x}+\frac{1}{x^{2}\sqrt{x}}$?.

If so, use the exponent laws to simplify before using product or quotient rules.

It simplifies to $\displaystyle x^{\frac{3}{2}}+x^{\frac{-5}{2}}$

In this case,no product or quotient rules needed.

$\displaystyle xsin(x)+\frac{sin(x)}{1+cos(x)}=xsin(x)+tan(\frac{x}{2})$

Use the product rule on the xsin(x) and the chain rule on the tan(x/2).

Remember, the derivative of tan(x) is $\displaystyle sec^{2}(x)$

 

[BigGlenntheHeavy]

First one:

$\displaystyle f(x) \ = \ (x)(x^{1/2})+\frac{1}{(x^{2})(x^{1/2})} \ = \ x^{3/2}+\frac{1}{x^{5/2}} \ = \ \frac{x^{4}+1}{x^{5/2}}$

$\displaystyle D_x\bigg[\frac{x^{4}+1}{x^{5/2}}\bigg] \ = \ \frac{x^{5/2}(4x^{3})-(x^{4}+1)(5/2)x^{3/2}}{x^{5}}$

$\displaystyle = \ \frac{4x^{11/2}- \frac{5x^{3/2}(x^{4}+1)}{2}} {x^{5}} \ = \ \frac{8x^{11/2}-5x^{11/2}-5x^{3/2}}{2x^{5}}$

$\displaystyle = \ \frac{3x^{11/2}-5x^{3/2}}{2x^{5}} \ = \ \frac{x^{3/2}(3x^{4}-5)}{2x^{5}} \ = \ \frac{3x^{4}-5}{2x^{7/2}}$

Second one:

$\displaystyle f(x) \ = \ xsin(x)+\frac{sin(x)}{1+cos(x)}$

$\displaystyle D_x\bigg[xsin(x)+\frac{sin(x)}{1+cos(x)}\bigg] \ = \ sin(x)+xcos(x)+\frac{(1+cos(x))cos(x)-sin(x)(-sin(x))}{(1+cos(x))^{2}}$

$\displaystyle = \ sin(x)+xcos(x)+\frac{cos(x)+cos^{2}(x)+sin^{2}(x)}{(1+cos(x))^{2}} \ = \ sin(x)+xcos(x)+\frac{1+cos(x)}{(1+cos(x))^{2}}$

$\displaystyle = \ sin(x)+xcos(x)+\frac{1}{1+cos(x)} \ = \ \frac{sin(x)+sin(x)cos(x)+xcos(x)+xcos^{2}(x)+1}{1+cos(x)}$

$\displaystyle Hence, \ f \ ' \ (x) \ = \ \frac{xcos^{2}(x)+cos(x)[x+sin(x)]+sin(x)+1}{1+cos(x)}, \ f \ ' \ (\pi/2) \ = \ 2$