Question regarding derivatives/ chain rule

[Danmann]

Hi all! I’m new to the forum and this is my first post. Thanks in advance for the help.

So the problem is y=cosx/sin^2x. Find the derivative. If I look at the problem as cosx/(sinx)^2 and differentiate using the quotient rule and chain rule, I get -sin^2x-2cos^2x/sin^3x.

But unless I’m missing something, cosx/sin^2x can also be written as (cosx/sinx)(1/sinx)= cotxcscx. When I differentiate this using only the product rule I get -cot^2xcscx-csc^3x.

Maybe I’m not seeing how these two answers are equivalent or have made a mistake or wrong assumption somewhere else. Not sure. Thanks for any clarification!

 

[willyengland]

Your assumption is correct, and your solutions are both correct.
If you plot them you will see that they agree.

 

[pka]

Hi all! I’m new to the forum and this is my first post. Thanks in advance for the help.

So the problem is y=cosx/sin^2x. Find the derivative. If I look at the problem as cosx/(sinx)^2 and differentiate using the quotient rule and chain rule, I get -sin^2x-2cos^2x/sin^3x.

But unless I’m missing something, cosx/sin^2x can also be written as (cosx/sinx)(1/sinx)= cotxcscx. When I differentiate this using only the product rule I get -cot^2xcscx-csc^3x.

Maybe I’m not seeing how these two answers are equivalent or have made a mistake or wrong assumption somewhere else. Not sure. Thanks for any clarification!

Use the web to check yourself.

 

[Dr.Peterson]

So the problem is y=cosx/sin^2x. Find the derivative. If I look at the problem as cosx/(sinx)^2 and differentiate using the quotient rule and chain rule, I get -sin^2x-2cos^2x/sin^3x.

But unless I’m missing something, cosx/sin^2x can also be written as (cosx/sinx)(1/sinx)= cotxcscx. When I differentiate this using only the product rule I get -cot^2xcscx-csc^3x.

Maybe I’m not seeing how these two answers are equivalent or have made a mistake or wrong assumption somewhere else. Not sure. Thanks for any clarification!

The interesting part is to try showing that the two are equivalent. You'll need to write both using correct parenthesization, so we can be sure what you mean.

I'd start by writing the second in terms of sine and cosine only, and then try to make it more like the first.

 

[Deleted member 4993]

Hi all! I’m new to the forum and this is my first post. Thanks in advance for the help.

So the problem is y=cosx/sin^2x. Find the derivative. If I look at the problem as cosx/(sinx)^2 and differentiate using the quotient rule and chain rule, I get (-sin^2x-2cos^2x)/sin^3x.

But unless I’m missing something, cosx/sin^2x can also be written as (cosx/sinx)(1/sinx)= cotxcscx. When I differentiate this using only the product rule I get -cot^2x * cscx - csc^3x.

Maybe I’m not seeing how these two answers are equivalent or have made a mistake or wrong assumption somewhere else. Not sure. Thanks for any clarification!

(-sin^2x - 2cos^2x) / sin^3x = -(1 + cos^2x) / sin^3x = -1 / sin^3x - cos^2x / sin^3x = - csc^3x - cot^2x * cscx

Those are same (after correcting for the parentheses).

 

[Danmann]

I appreciate the replies. I initially made a mistake entering the two solutions into my graphing app. I now see that they are indeed equivalent. And thanks Subhotosh for clarifying how to make the two solutions identical.