d/dx( ln |sin(pi/x)| ) = ?
Thanks.
If those are absolute value signs, the derivative will not exist when sin (pi/x) = 0, because of the sign change that occurs there. Assume sin (pi/x) > 0
Let u(x) = pi/x and v(x) = sin x, and use the chain rule.
d/dx ln v(u(x))=
d/dv dv/du du/dx
= -[1/(sin (pi/x])*cos x*(pi/x^2)
That assumes sin (pi/x) is positive. Change the sign if it is negative
.
To summarize:
To find the derivative of ln |sin(pi/x)|, we have to consider the absolute value sign. The derivative will not exist when sin(pi/x) = 0 because of the sign change that occurs there.
Assuming sin(pi/x) > 0, we can proceed by letting u(x) = pi/x and v(x) = sin(x), and using the chain rule.
The derivative can be computed as:
d/dx ln |sin(pi/x)| = d/dv dv/du du/dx
= -[1/(sin(pi/x))] * cos(x) * (pi/x^2)
It is important to note that the sign of the derivative will change if sin(pi/x) is negative.
To find the derivative of ln |sin(pi/x)|, we need to use the chain rule.
Assuming sin(pi/x) is positive, we can express the function as ln(sin(pi/x)).
Let's introduce two new functions for ease of computation:
u(x) = pi/x and v(x) = sin(u(x)).
Now, we will differentiate the function ln(v(u(x))) using the chain rule.
The chain rule states that the derivative of ln(f(x)) is equal to the derivative of f(x) divided by f(x).
So, d/dx ln(v(u(x))) = (dv/du * du/dx) / v(u(x)).
Now, let's calculate each part individually:
dv/du represents the derivative of v with respect to u. This is simply the derivative of sin(u), which is cos(u).
So, dv/du = cos(u).
du/dx represents the derivative of u with respect to x. To find this, we differentiate u(x) using the quotient rule.
du/dx = -π/x^2
Putting it all together, we have:
d/dx ln(v(u(x))) = (cos(u) * (-π/x^2)) / v(u(x)).
Since we assumed sin(pi/x) is positive, the value of v(u(x)) is equal to sin(pi/x).
Therefore, the final derivative is:
d/dx ln |sin(pi/x)| = -[1/(sin(pi/x))] * cos(pi/x) * (π/x^2).
If sin(pi/x) is negative, we need to change the sign of the derivative.