Implicit Differentiation Problem - Check my work?

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SUMMARY

The discussion focuses on implicit differentiation, specifically solving the equation xcos(y) + ycos(x) = 1. The user correctly derives dy/dx as (ysinx - cosy)/(-xsiny + cos x) after applying the product rule and simplifying the equation. A second problem involving the equation squareroot(xy) = 1 + (x^2)y is also addressed, with the correct differentiation yielding dy/dx = (2xy - y)/(x - x^2). The importance of careful notation and application of differentiation rules is emphasized throughout the discussion.

PREREQUISITES
  • Understanding of implicit differentiation
  • Familiarity with trigonometric functions and their derivatives
  • Knowledge of the product rule in calculus
  • Ability to manipulate algebraic expressions
NEXT STEPS
  • Study implicit differentiation techniques in calculus
  • Review the product rule and its applications in differentiation
  • Explore trigonometric derivatives and their implications
  • Practice solving complex implicit differentiation problems
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Students and educators in calculus, particularly those focusing on implicit differentiation and trigonometric functions. This discussion is beneficial for anyone looking to enhance their understanding of differentiation techniques.

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Implicit Differentiation Problem -- Check my work?

I've worked it -- can someone just check my work?

Problem:

xcosy+ycos=1

My work:

[x (d/x)cosy + cosy (d/dx)x] + [y (d/dx)cosx + cosx (d/dx)y] = (d/dx) 1

-xsiny (dy/dx) + cos y - ysinx + cos x (dy/dx) = 0

-xsiny (dy/dx) + cos y = ysinx - cosy

dy/dx = (ysinx - cosy)/(-xsiny + cos x)


Meanwhile, could someone help me with this one...

squareroot (xy) = 1+(x^2)y
 
Last edited:
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the first one looks good, as for the second:



(2)[tex]xy=1+x^2y[/tex]

You are differentiating WRT y, which means that the derivative of x is 1, but the derivative of y is dy/dx. Make sure you use the product rule:

[tex]x\frac{dy}{dx}+y=x^2 \frac{dy}{dx}+2xy[/tex]

[tex]x\frac{dy}{dx}-x^2 \frac{dy}{dx}=2xy-y[/tex]

[tex]\frac{dy}{dx}(x-x^2)=2xy-y[/tex]

[tex]\frac{dy}{dx}=\frac{2xy-y}{x-x^2}[/tex]

~Josh
 
-xsiny (dy/dx) + cos y = ysinx - cosy

dy/dx = (ysinx - cosy)/(-xsiny + cos x)

You miswrote the first line above but obviously that was a typo since you got it right in the end. If I were your teacher I would prefer to see one more line:
-xsiny (dy/dx) + cos y = ysinx - cos x(dy/dx)

cos x (dy/dx)- x sin y (dy/dx)= (cos x- x sin y)(dy/dx)= y sin x- cos y

dy/dx = (ysinx - cosy)/(-xsiny + cos x)
 

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