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Mathematics 9 Online
OpenStudy (anonymous):

calc

OpenStudy (anonymous):

OpenStudy (anonymous):

can someone help me with these 2 questions

hartnn (hartnn):

which step are you stuck ?

OpenStudy (anonymous):

For one, I was trying to follow the solution sheet, but I couldnt get it

OpenStudy (anonymous):

they got rid of tan but I dont see how they did that

hartnn (hartnn):

can you show us couple of steps...or the last step that you are sure is correct ?

OpenStudy (anonymous):

I started to understand implicit but kinda got stuck again

hartnn (hartnn):

derivative of tan^-1 x is ?

OpenStudy (anonymous):

okay hold on anddd um not sure of inverse

OpenStudy (anonymous):

1/sec^2x?

OpenStudy (anonymous):

oh 1/1+x^2?

hartnn (hartnn):

yes

hartnn (hartnn):

so , applying chain rule derivative of tan^-1 u will be 1/(1+u^2) du/dx

OpenStudy (anonymous):

Okay lemme try and I will show you what I have after ?

hartnn (hartnn):

u = 2x^2 y

hartnn (hartnn):

sure,

OpenStudy (anonymous):

okay gimme 5 mins

OpenStudy (anonymous):

for the x+4xy^2 part

OpenStudy (anonymous):

No stuck on that part, Im a bit confused

OpenStudy (anonymous):

on what to do there

hartnn (hartnn):

derivative of x is just 1 for 4xy^2 you'll need product + chain rule

OpenStudy (anonymous):

okay so then, 8x dydx + 2y dydx?

OpenStudy (anonymous):

no I think I did it wrong

hartnn (hartnn):

4 (d/dx (x) y^2 + x d/dx (y^2)) = 4 ( y^2 + x [2y dy/dx] )

hartnn (hartnn):

see if you get that application of product rule ^^

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