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The derivative of sin(x^1/x)?

User Www
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1 Answer

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Finding the derivative of sin(x^1/x) can be quite tricky. We can use the chain rule to approach this problem.

Let's define y = x^1/x. Then, we can write sin(x^1/x) as sin(y).

Now, let's find the derivative of y with respect to x:

y = x^(1/x)
ln(y) = (1/x)*ln(x)
Differentiating both sides with respect to x, we get:
(1/y)*(dy/dx) = (-1/x^2)*ln(x) + (1/x^2)
(dy/dx) = y*(-ln(x)/x^2 + 1/x^2)
(dy/dx) = x^(1/x)*(-ln(x)/x^2 + 1/x^2)

Next, using the chain rule:

d/dx(sin(y)) = cos(y)*(dy/dx)
Substituting dy/dx from the previous step:

d/dx(sin(x^1/x)) = cos(x^1/x)*x^(1/x)*(-ln(x)/x^2 + 1/x^2)

Therefore, the derivative of sin(x^1/x) is cos(x^1/x)*x^(1/x)*(-ln(x)/x^2 + 1/x^2).
User Gucal
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