Question

First derivative cot(x)cos(x).
The correct answer is -cosx-cotxcscx, but I don't know how.

Answer 1

how would you start ??

Answer 2

\[\large\rm \cot(x) \cdot \cos(x)\]
there are two functions so apply the product rule ?\[\large\rm y=f(x) \cdot g(x) ~~~~~~~~~y'= f'(x) \cdot g(x) + f(x) \cdot g'(x)\]

Answer 3

Yes, when I use that method I get -csc^2(x)cos(x)-cot(x)sin(x)

Answer 4

yep looks good now rewrite cot and csc^2 in terms of cos ., sin :)

Answer 5

\[\large\rm -\color{reD}{\csc^2(x)}\cos(x)-\color{blue}{\cot(x)}\sin(x)\]
\[\large\rm -\color{red}{\frac{ 1 }{ \sin^2x }} \cos(x) - \color{blue}{\frac{\cos x}{\sin x}} \cdot \sin(x)\]
sin^2x is same as (sinx times sin x)
so \[\large\rm -\color{red}{\frac{ 1 }{ sinx \cdot \sin x }} \cos(x) - \color{blue}{\frac{\cos x}{\sin x}} \cdot \sin(x)\]
now simpify

Answer 6

let me know if you have question

Answer 7

I'm not entirely sure how to ask this question, but I'll try.
Why can the sin(x) on right side of the subtraction sign simplify the sin(x) on the left side of the subtraction sign? I thought it couldn't do that.

Answer 8

@Nnesha

Answer 9

http://prnt.sc/aingsr
are you talking about these sin ratios ?

Answer 10

@Nnesha yes I am

Answer 11

ohh yea we can't do that
bec 1/sin^2 is a different term
just like in this example
we can't just cancel out y . first we need to find common denominator
\[\large\rm \frac{1}{y}- y\]

Answer 12

\[\large\rm -\color{red}{\frac{ 1 }{ sinx \cdot \sin x }} \cos(x) - \color{blue}{\frac{\cos x}{\cancel{\sin x}}} \cdot \cancel{\sin(x)}\]

you can cancel out sinx in 2nd term bec its a multiplication
i don't know if that make sense .. hm

Answer 13

I just realized something, the way you're solving the problem doesn't give me the correct answer either

Answer 14

@Nnesha I figured out, you weren't wrong I was just visualizing my answer incorrectly. Thank you for your help!

Answer 15

aw glad to hear that
if i split this fraction into two fractions then it would be easy to understand
\[\large\rm -\color{red}{\frac{ 1 }{ sinx }\cdot \frac{1}{\sin x }} \cdot \cos(x) - \color{blue}{\frac{\cos x}{\cancel{\sin x}}} \cdot \cancel{\sin(x)}\]
\[\large\rm -\color{red}{\frac{ 1 }{ sinx }\cdot \frac{\color{black}{cos x }}{\sin x }} - \color{blue}{\frac{\cos x}{\cancel{\sin x}}} \cdot \cancel{\sin(x)}\]
:)

Answer 16

& my pleasure :)