Trying to find dy/dx of a trig function # 2

  • Thread starter jtt
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  • #1
jtt
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Homework Statement


find dy/dx


Homework Equations


x+tanxy=0


The Attempt at a Solution


d/dy(x+tanxy)

x+sec^2(xy)((1)(dy/dx))+(1)(tanxy)=0
dy/dx(sec^2(xy)+x+tanxy=0
-x-tanxy -x-tanxy
dy/dx(sec^2(xy)/(sec^2(xy)=(-x-tanxy)/(sec^2(xy))

dy/dx=(-x-tanxy)/(sec^2xy)
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement


find dy/dx


Homework Equations


x+tanxy=0

You can begin by stating the problem unambiguously. Are you trying to differentiate

x + tan(xy) = 0 or x + ytan(x)=0. The point is that as it is written we can't tell whether the y is inside or outside that tangent function. Parentheses are necessary!

The Attempt at a Solution


d/dy(x+tanxy)

Why are you writing d/dy when you are differentiating with respect to x?
 
  • #3
jtt
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trying to differentiate x+tan(xy)

i got dy/dx when i took the derivative of y in tan(xy)
 
  • #4
LCKurtz
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Homework Statement


find dy/dx


Homework Equations


x+tanxy=0


The Attempt at a Solution


d/dy(x+tanxy)

You mean d/dx(x + tan(xy))

x+sec^2(xy)((1)(dy/dx))+(1)(tanxy)=0
Is the derivative of x equal to x??

And what I highlighted in red should be the derivative of the (xy) which is the argument of the tangent function, or the "inside". There should be no tan(xy) in that.
 

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