CA Foundation Exam > CA Foundation Questions > 2^log2^x then dy/dx?
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2^log2^x then dy/dx?
Most Upvoted Answer
2^log2^x then dy/dx?
Solution:
Using Chain Rule
We can use chain rule to differentiate the given expression.
Let y = 2^(log2^x)
Taking natural logarithm on both sides, we get
ln y = ln(2^(log2^x))
Applying exponent rule of logarithm, we get
ln y = log2^x * ln 2
Differentiating both sides with respect to x, we get
1/y * dy/dx = (1/ln 2) * (1/x) * (d/dx) [log2^x]
Using chain rule, we get
dy/dx = y * (1/ln 2) * (1/x) * 2^(log2^x) * (d/dx) [log2^x]
Simplifying the expression
Now we can simplify the expression by substituting the values.
y = 2^(log2^x)
= x (since 2^log2^x = x)
Therefore,
dy/dx = x * (1/ln 2) * (1/x) * 2^(log2^x) * (d/dx) [log2^x]
= (1/ln 2) * 2^x
Final Answer
Therefore, the derivative of 2^log2^x with respect to x is (1/ln 2) * 2^x.
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Question Description
2^log2^x 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 2^log2^x then dy/dx? covers all topics & solutions for CA Foundation 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2^log2^x then dy/dx?.
2^log2^x 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 2^log2^x then dy/dx? covers all topics & solutions for CA Foundation 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2^log2^x then dy/dx?.
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