finding the trignometric derivative

[catalle]

so I did some steps for this problem and got stuck halfway down, please help

find dy/dx
y=1+x-cos(x)
= [1-cos(x)] + x
= sin(x) + x

[sin (x+h) + (x+h) - sin (x) - x]/h --- definition of derivative

[sin(x)*cos(h) + cos(x)*sin(h) + x + h - sin(x)-x] / h --- angle sum rule?

[sin(x)*cos(h) + cos(x)*sin(h) + h - sin(x)] / h --- eliminate Xs

do I eliminate the Hs now as well? but then there wouldn't be an h on the denominator
do I plug in 0 for h now or what?

 

[galactus]

$\displaystyle 1-cos(x)\neq sin(x)$

Unless you have a typo.

Do you HAVE to use the def. of a derivative approach?.

 

[catalle]

yeah I think so, then the product/quotient rule

 

[galactus]

You are correct about using the addition rule for cos.

$\displaystyle \lim_{h\to 0}\frac{1+(x+h)-cos(x+h)-1-x+cos(x)}{h}$

$\displaystyle \lim_{h\to 0}\frac{h-cos(x)cos(h)+sin(x)sin(h))+cos(x)}{h}$

Factor out a cos(x) and a sin(x). Then, notice the famous limits $\displaystyle \frac{cos(h)-1}{h}=0, \;\ \frac{sin(x)}{x}=1$

$\displaystyle \lim_{h\to 0}1-cos(x)\left(\frac{cos(h)-1}{h}\right)+sin(x)\left(\frac{sin(x)}{x}\right)$