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Y=x^log(logx) then dy/dx=?
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Y=x^log(logx) then dy/dx=?
Given Function: y = x^(log(log(x)))

To find: dy/dx

Solution:

Step 1: Take the natural logarithm (ln) of both sides of the equation to simplify the function.

ln(y) = ln(x^(log(log(x))))

Step 2: Apply the logarithmic properties to simplify the expression.

ln(y) = log(log(x)) * ln(x)

Step 3: Differentiate both sides of the equation implicitly with respect to x.

d/dx(ln(y)) = d/dx(log(log(x)) * ln(x))

Using the chain rule:

1/y * dy/dx = d/dx(log(log(x))) * ln(x) + log(log(x)) * d/dx(ln(x))

Step 4: Simplify the derivatives.

1/y * dy/dx = (1/log(x)) * (1/x) * ln(x) + log(log(x)) * (1/x)

Step 5: Multiply both sides of the equation by y to isolate dy/dx.

dy/dx = y * [(1/log(x)) * (1/x) * ln(x) + log(log(x)) * (1/x)]

Step 6: Substitute the value of y from the given equation.

dy/dx = x^(log(log(x))) * [(1/log(x)) * (1/x) * ln(x) + log(log(x)) * (1/x)]

Step 7: Simplify the expression further if possible.

The above expression is the final derivative of y with respect to x.

Summary:

The derivative of y = x^(log(log(x))) with respect to x is dy/dx = x^(log(log(x))) * [(1/log(x)) * (1/x) * ln(x) + log(log(x)) * (1/x)].
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Y=x^log(logx) then dy/dx=?
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Y=x^log(logx) then dy/dx=? for CA Foundation 2025 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about Y=x^log(logx) then dy/dx=? covers all topics & solutions for CA Foundation 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Y=x^log(logx) then dy/dx=?.
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