def. of deriv

**kwivo** · October 7th 2007, 12:36 PM

Use the definition of derivative to find d/dx cos(x).

**Jhevon** · October 7th 2007, 12:42 PM

Originally Posted by kwivo

Use the definition of derivative to find d/dx cos(x).

by the limit definition of the derivative:

$\frac d{dx}f(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}h$

thus, $\frac d{dx} \cos(x) = \lim_{h \to 0}\frac {\cos (x + h) - \cos x }h$

now, continue (begin by using the addition formula for cosine to expand $\cos (x + h)$ in the numerator

**galactus** · October 7th 2007, 12:48 PM

You can use the addtion formula for cos.

$\frac{d}{dx}[cos(x)]=\lim_{h\rightarrow{0}}\frac{cos(x+h)-cos(x)}{h}$

= $\lim_{h\rightarrow{0}}\frac{cos(x)cos(h)-sin(x)sin(h)-cos(x)}{h}$

= $\lim_{h\rightarrow{0}}\left[cos(x)\left(\frac{cos(h)-1}{h}\right)-sin(x)\left(\frac{sin(h)}{h}\right)\right]$

Can you finish?.

**kwivo** · October 7th 2007, 12:57 PM

Ok, I'm not sure how to get rid of the cos(h)-1/h, but the sin(h)/h is equal to 1 right?

**Jhevon** · October 7th 2007, 12:58 PM

Originally Posted by kwivo

Ok, I'm not sure how to get rid of the cos(h)-1/h, but the sin(h)/h is equal to 1 right?

$\lim_{x \to 0} \frac {\cos x - 1}x$ happens to be a special limit. look it up. ... nevermind, it's equal to zero.

yes, $\lim_{h \to 0} \frac {\sin h}h = 1$

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