Derivative of \( \displaystyle \left(2 x + 1\right)^{2 x} \)

Problem 2.366 · medium

1. \[ \frac{d}{d x} \left(2 x + 1\right)^{2 x} \]
   derivativeStart with the derivative of the function.✓ Proved
2. \[ = \frac{d}{d x} e^{2 x \ln{\left(2 x + 1 \right)}} \]
   rewriteRewrite the base and exponent using the exponential and logarithm.≈ Checked numerically
3. \[ = e^{2 x \ln{\left(2 x + 1 \right)}} \frac{d}{d x} 2 x \ln{\left(2 x + 1 \right)} \]
   chainApply the chain rule for the exponential function.✓ Proved
4. \[ = \left(2 x \frac{d}{d x} \ln{\left(2 x + 1 \right)} + \ln{\left(2 x + 1 \right)} \frac{d}{d x} 2 x\right) e^{2 x \ln{\left(2 x + 1 \right)}} \]
   productApply the product rule to the inner expression.✓ Proved
5. \[ = \left(2 x \frac{d}{d x} \ln{\left(2 x + 1 \right)} + 2 \ln{\left(2 x + 1 \right)}\right) e^{2 x \ln{\left(2 x + 1 \right)}} \]
   derivativeDifferentiate 2*x.✓ Proved
6. \[ = \left(\frac{2 x \frac{d}{d x} \left(2 x + 1\right)}{2 x + 1} + 2 \ln{\left(2 x + 1 \right)}\right) e^{2 x \ln{\left(2 x + 1 \right)}} \]
   derivativeDifferentiate the logarithm using the chain rule.✓ Proved
7. \[ = \left(\frac{4 x}{2 x + 1} + 2 \ln{\left(2 x + 1 \right)}\right) e^{2 x \ln{\left(2 x + 1 \right)}} \]
   derivative algebraDifferentiate 2*x + 1. Simplify the expression inside the parentheses.✓ Proved
8. \[ = \left(2 x + 1\right)^{2 x} \left(\frac{4 x}{2 x + 1} + 2 \ln{\left(2 x + 1 \right)}\right) \]
   simplifyConvert the exponential form back to the original power form.≈ Checked numerically