Implicit Differentiation w/ trig functions check

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Homework Help Overview

The problem involves implicit differentiation of the equation y*sin(x²) = 5, specifically determining dy/dx. The subject area is calculus, focusing on differentiation techniques involving trigonometric functions.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the manipulation of terms in the differentiation process, questioning the transition of terms between the numerator and denominator. There are attempts to clarify the steps taken in isolating dy/dx and concerns about the legality of certain algebraic manipulations.

Discussion Status

The discussion is active, with participants providing feedback on each other's attempts. Some guidance has been offered regarding isolating dy/dx correctly, and there is a recognition of improvements in the latest attempts at the solution.

Contextual Notes

Participants are navigating the complexities of implicit differentiation and the specific algebraic steps involved, indicating a focus on understanding the process rather than arriving at a final answer.

DollarBill
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Homework Statement


Determine dy/dx when

y*sin(x2)=5

The Attempt at a Solution


y*2xcos(x2) dx/dx + sin(x2)*1 dy/dx = 0

2xy cos(x2)*dy/dx = -sin(x2)

dy/dx = -sin(x2) / 2xy cos(x2)

dy/dx = -2xy tan(x2)
 
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How did 2xy magically pop from the denominator to the numerator?
 
Would it be +2xy tan(x2)?
 
y*2xcos(x2) dx/dx + sin(x2)*1 dy/dx = 0

2xy cos(x2)*dy/dx = -sin(x2)

What I notice is that you separated sin(x^2)dy/dx by subtraction..which is obviously "illegal"

Try isolating the term with dy/dx in it and then try to isolate just the dy/dx
 
I didn't even notice I did that

Working it out again:

y*2xcos(x2) dx/dx + sin(x2)dy/dx = 0

sin(x2)dy/dx = -2xycos(x2)

dy/dx = -2xycos(x2) / sin(x2)

dy/dx=-2xy*Cotx2
 
That looks better.
 

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