Question

find the derivative of the function f (x)=
$(1 - \sin(x)) \div (1 + \cos(x) )$
please solve step by step ​

Answer 1

Answer:

The trigonometric functions \sin(x)sin(x)sine, left parenthesis, x, right parenthesis and \cos(x)cos(x)cosine, left parenthesis, x, right parenthesis play a significant role in calculus. These are their derivatives:

\begin{aligned} \dfrac{d}{dx}[\sin(x)]&=\cos(x) \\\\ \dfrac{d}{dx}[\cos(x)]&=-\sin(x) \end{aligned}  

dx

d

​  

[sin(x)]

dx

d

​  

[cos(x)]

​  

 

=cos(x)

=−sin(x)

​  

 

The AP Calculus course doesn't require knowing the proofs of these derivatives, but we believe that as long as a proof is accessible, there's always something to learn from it. In general, it's always good to require some kind of proof or justification for the theorems you learn.

First, we would like to find two tricky limits that are used in our proof.

1. \displaystyle\lim_{x\to 0}\dfrac{\sin(x)}{x}=1  

x→0

lim

​  

 

x

sin(x)

​  

=1limit, start subscript, x, \to, 0, end subscript, start fraction, sine, left parenthesis, x, right parenthesis, divided by, x, end fraction, equals, 1

2. \displaystyle\lim_{x\to 0}\dfrac{1-\cos(x)}{x}=0  

x→0

lim

​  

 

x

1−cos(x)

​  

=0limit, start subscript, x, \to, 0, end subscript, start fraction, 1, minus, cosine, left parenthesis, x, right parenthesis, divided by, x, end fraction, equals, 0

Now we are ready to prove that the derivative of \sin(x)sin(x)sine, left parenthesis, x, right parenthesis is \cos(x)cos(x)cosine, left parenthesis, x, right parenthesis.

Finally, we can use the fact that the derivative of \sin(x)sin(x)sine, left parenthesis, x, right parenthesis is \cos(x)cos(x)cosine, left parenthesis, x, right parenthesis to show that the derivative of \cos(x)cos(x)cosine, left parenthesis, x, right parenthesis is -\sin(x)−sin(x)minus, sine, left parenthesis, x, right parenthesis.

Step-by-step explanation: there you go