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Mathematics 16 Online
OpenStudy (anonymous):

Find derivative of cos(xy)=xy. Please show all steps

OpenStudy (anonymous):

Is this partial derivatives?

OpenStudy (anonymous):

Implicit relations

OpenStudy (anonymous):

What calculus lvl though? 1 or 3?

OpenStudy (anonymous):

Level 1. This is ap calculus ab

OpenStudy (anonymous):

Oh sorry i mean i remember how to do it in calc 3 but not calc one

OpenStudy (anonymous):

Oh I see. But how'd uoi like maybe solve it still though with calc 3? Wouldn't they be like the same way of solving maybe?

OpenStudy (anonymous):

Oh I see. But how'd uoi like maybe solve it still though with calc 3? Wouldn't they be like the same way of solving maybe?

OpenStudy (anonymous):

not quite since i would use partial derivatives to solve it

OpenStudy (anonymous):

and i guess in a sense i could try to explain how to do it to you right now since its a little simple

OpenStudy (anonymous):

But all it just is, is taking the partial derivative of x while keeping y constant, and then taking the partial derivative of y while keeping x constant

OpenStudy (anonymous):

and when i say constant i mean making certain variable a constant, like a constant number

OpenStudy (anonymous):

like 1 or 2 or 85748457 for that matter since we know that the derivative of a constant is always zero

OpenStudy (anonymous):

Alright

OpenStudy (anonymous):

Hey man can you help me with some other problems also?

OpenStudy (anonymous):

I can try but its been awhile since i have done calc 1

OpenStudy (anonymous):

Alright. Find derivative of y=tan^-1(x/2)

OpenStudy (anonymous):

is this implicit defferentiation again?

OpenStudy (anonymous):

or did you just want me to find the derivative of the tan inverse part?

OpenStudy (anonymous):

No. Just regular derivative

OpenStudy (anonymous):

alright well we know right from the bat that this is going to have to use the chain rule at the beginning

OpenStudy (anonymous):

Since we need to take the derivative of the inside * the derivative of the outside

OpenStudy (anonymous):

Alright

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