Mathematics
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OpenStudy (anonymous):
calc
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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
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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?
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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
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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
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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 ^^