How to differentiate x^(cosx) = y^(sinx) with respect to x

In summary, the conversation is about differentiating x^(cosx) = y^(sinx) with respect to x using natural logs. The attempt at a solution involves using the differentiation rule and simplifying the equation using parentheses.
  • #1
styxrihocc
10
0

Homework Statement



Differentiate x^(cosx) = y ^(sinx) with respect to x

Homework Equations


The Attempt at a Solution


I tried using natural logs but I am not sure if its correct, if it's wrong please point me to the right direction, thanks

x^(cosx) = y^(sinx)
ln x^(cosx) = ln y ^(sinx)
ln x (cosx) = ln y (sinx)
cosx/x - sinx lnx = cosx lny +sinx/y (dy/dx)
cosx/x - sinx lnx - cosx lny = sinx/y (dy/dx)
(cosx/x - sinx lnx - cosx lny) / (sinx/y) = dy/dx
 
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  • #2
styxrihocc said:

Homework Statement



Differentiate x^(cosx) = y ^(sinx) with respect to x

Homework Equations





The Attempt at a Solution


I tried using natural logs but I am not sure if its correct, if it's wrong please point me to the right direction, thanks

x^(cosx) = y^(sinx)
ln x^(cosx) = ln y ^(sinx)
ln x (cosx) = ln y (sinx)
cosx/x - sinx lnx = cosx lny +sinx/y (dy/dx)
cosx/x - sinx lnx - cosx lny = sinx/y (dy/dx)
(cosx/x - sinx lnx - cosx lny) / (sinx/y) = dy/dx

Your work is correct. You could be slightly less ambiguous with parentheses though.
 
  • #3
Thanks.
 

1. What is the first step in differentiating x^(cosx) = y^(sinx) with respect to x?

The first step is to take the natural logarithm of both sides of the equation, which will help to simplify the problem.

2. Why do we need to take the natural logarithm?

Taking the natural logarithm of both sides allows us to use the logarithmic rule for differentiating exponents, which states that d/dx(ln(u)) = u'/u.

3. How do we differentiate x^(cosx) and y^(sinx)?

To differentiate x^(cosx), we use the chain rule, where we first take the derivative of the outer function, cosx, and then multiply it by the derivative of the inner function, x. Similarly, for y^(sinx), we take the derivative of the outer function, sinx, and multiply it by the derivative of the inner function, y.

4. What is the derivative of cosx and sinx?

The derivative of cosx is -sinx, and the derivative of sinx is cosx.

5. How do we differentiate the left side of the equation?

Using the chain rule, we differentiate x^(cosx) to get -x^(cosx) * ln(x) * sinx. Similarly, we differentiate y^(sinx) to get -y^(sinx) * ln(y) * cosx.

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