How Do You Differentiate sin, cos, and tan Functions?

[EIRE2003]

What are the rules for differentiating tan, sin & cos?

I know cos = -sin

tan = sin/cos?

 

[whozum]

Be careful of what you type.

You can derive sin's derivative (say that five times fast) using the limit definition, from there a simple Taylor series expansion will get you the other two, or knowing sinx' = cosx and cosx' = -sinx, you can use the quotient rule to find tanx.

 

[PhysicsinCalifornia]

Yea, just like what Whozum said...

if $$f(x) = \sin \theta$$,$$f'(x) = \cos \theta$$
$$f(x) = \cos \theta$$, $$f'(x) = -\sin \theta$$
$$f(x) = \tan \theta$$, $$f'(x) = \sec^2 \theta$$
$$f(x) = \csc \theta$$, $$f'(x) = -\csc\theta \cot\theta$$
$$f(x) = \sec \theta$$, $$f'(x) = \sec\theta \tan\theta$$
$$f(x)= \cot\theta$$, $$f'(x) = -\csc^2\theta$$

Try finding the derivative of the inverse trig functions

 

[Jameson]

You are correct $$\frac{d}{dx}\cos{x} = -\sin{x}$$

,and I assume you know that $$\frac{d}{dx}\sin{x} = \cos{x}$$.

Using these two definitions, use the quotient rule to find the derivative of tanx as Whozum said above.

$$\frac{d}{dx}\frac{u}{v} = \frac{vu' - uv'}{v^2}$$

Express tangent as $$\frac{\sin{x}}{\cos{x}}$$ and see what you get using the rule above.

Jameson