Find derivatives of following functions . y=a^x(agt 0,aneq 1) y=x^x

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Question

Find derivatives of following functions . y=a^x(agt 0,aneq 1) y=x^x

Answer 1

1. The derivative of \(y = a^x\) is \(y' = a^x \ln(a)\). 2. The derivative of \(y = x^x\) is \(y' = x^x (\ln(x) + 1)\).

Answer 2

## Step 1: To find the derivative of the function \(y = a^x\), where \(a > 0\) and \(a \neq 1\), we use the rule of differentiation for exponential functions. ### **The derivative of \(a^x\) with respect to \(x\) is \(a^x \ln(a)\)** This rule applies because \(a\) is a constant and \(x\) is the variable. ## Step 2: For the function \(y = x^x\), we use the chain rule and the property of logarithmic differentiation. This is because both the base and the exponent are variables. ## Step 3: We first transform the function by taking the natural logarithm of both sides, which gives us \(\ln(y) = \ln(x^x)\). ## Step 4: Simplifying the right side, we get \(\ln(y) = x \ln(x)\). ## Step 5: Next, we differentiate both sides with respect to \(x\) and use the product rule on the right side, which gives us \(\frac{1}{y} \cdot \frac{dy}{dx} = \ln(x) + x \cdot \frac{1}{x}\). ## Step 6: Solving for \(\frac{dy}{dx}\), we find \(\frac{dy}{dx} = y \cdot (\ln(x) + 1)\). ## Step 7: Finally, we substitute back \(y = x^x\), which gives us \(\frac{dy}{dx} = x^x (\ln(x) + 1)\).

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Find derivatives of following functions . y=a^x(agt 0,aneq 1) y=x^x

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