ABCD is a parallelogram.

Mathematics

1 answer:

Maslowich3 years ago

5 0

Answer:

It is

Step-by-step explanation:

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(a) Use the Quotient Rule to differentiate the function f(x)=tan(x)-1/sec(x). f'(x)=

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Answer:

(a) sin(x) + cos(x)

(b) sin(x) + cos(x)

Step-by-step explanation:

(a) The given function is:

$f(x) = \frac{tan(x)-1}{sec(x)}$

According to the quotient rule:

$f(x) = \frac{g(x)}{h(x)}\\f'(x)= \frac{g'(x)h(x)-g(x)h'(x)}{h(x)^2}$

Applying the quotient rule:

$f(x) = \frac{tan(x)-1}{sec(x)}\\f'(x)=\frac{sec^2(x)*sec(x)-(tan(x)-1)*sec(x)tan(x)}{sec(x)^2}\\f'(x)=\frac{sec^3(x)-sec(x)tan^2(x)+sec(x)tan(x)}{sec(x)^2}\\f'(x)=sec(x)+\frac{tan(x)-tan^2(x)}{sec(x)}\\ \frac{tan(x)}{sec(x)}=sin(x) \\f'(x)=sec(x)+sin(x)-sin(x)tan(x)\\$

This can be simplified to:

$f'(x)=sec(x)+sin(x)-sin(x)tan(x)\\f'(x) = \frac{1}{cos(x)}+sin(x)-\frac{sin^2(x)}{cos(x)}\\f'(x)=\frac{1+sin(x)cos(x)-sin^2(x)}{cos(x)}\\f'(x)=\frac{sin^2(x)+cos^2(x)+sin(x)cos(x)-sin^2(x)}{cos(x)}\\f'(x)=\frac{cos^2(x)+sin(x)cos(x)}{cos(x)}\\ f'(x)=sin(x) +cos(x)$