How do you differentiate #y=(1+x)^(1/x)#?

Question

Aurora Alvarado · Accepted Answer

# dy/dx = (1+x)^(1/x) {1/(x(1+x)) - ln(1+x)/x^2} #

We have: # y = (1+x)^(1/x) # First note that the function is not defined when #x=0#. We can take (natural) logarithms of both sides to get: # lny = ln{(1+x)^(1/x)} # # " " = 1/x \ ln(1+x) # # " " = ln(1+x)/x # Differentiating implicitly and applying the quotient rule and the chain rule gives: # 1/y dy/dx = { (x)(1/(1+x)) - (ln(1+x))(1) } / (x)^2 # # " " = { x/(1+x) - ln(1+x) } / (x)^2 # # " " = 1/(x(1+x)) - ln(1+x)/x^2 # And so: # dy/dx = y {1/(x(1+x)) - ln(1+x)/x^2} # # " " = (1+x)^(1/x) {1/(x(1+x)) - ln(1+x)/x^2} #

Elijah Abram · Answer

The answer is #=(1+x)^(1/x)(1/(x(1+x))-ln(1+x)/x^2)#

We diiferentiate using logs.

#y=(1+x)^(1/x)#

#ln(y)=ln((1+x)^(1/x))#

#lny=1/x*ln(1+x)#

Differentiating

#(lny)'=(1/x*ln(1+x))'#

#1/y*dy/dx=1/x*1/(1+x)-1/x^2*ln(1+x)#

#dy/dx=y(1/(x(1+x))-1/x^2ln(1+x))#

#dy/dx=(1+x)^(1/x)(1/(x(1+x))-ln(1+x)/x^2)#

Elijah Abram · Answer

undefined To differentiate $ y = (1 + x)^{\frac{1}{x}} $, we will use the logarithmic differentiation method.

1. Take the natural logarithm of both sides:
$$ \ln(y) = \ln\left((1 + x)^{\frac{1}{x}}\right) $$
$$ \ln(y) = \frac{1}{x} \ln(1 + x) $$

2. Differentiate implicitly with respect to $ x $:
$$ \frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{x} \ln(1 + x)\right) $$

3. Use the product rule and the chain rule to differentiate the right-hand side:
$$ \frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{x}\right) \cdot \ln(1 + x) + \frac{1}{x} \cdot \frac{d}{dx}(\ln(1 + x)) $$
$$ \frac{1}{y} \cdot \frac{dy}{dx} = -\frac{1}{x^2} \cdot \ln(1 + x) + \frac{1}{x} \cdot \frac{1}{1 + x} $$

4. Simplify the expression:
$$ \frac{1}{y} \cdot \frac{dy}{dx} = -\frac{\ln(1 + x)}{x^2} + \frac{1}{x(1 + x)} $$

5. Now, find $ \frac{dy}{dx} $ by multiplying both sides by $ y $ (which is $ (1 + x)^{\frac{1}{x}} $):
$$ \frac{dy}{dx} = y \left(-\frac{\ln(1 + x)}{x^2} + \frac{1}{x(1 + x)}\right) $$
$$ \frac{dy}{dx} = (1 + x)^{\frac{1}{x}} \left(-\frac{\ln(1 + x)}{x^2} + \frac{1}{x(1 + x)}\right) $$

Therefore, the derivative of $ y = (1 + x)^{\frac{1}{x}} $ is $ (1 + x)^{\frac{1}{x}} \left(-\frac{\ln(1 + x)}{x^2} + \frac{1}{x(1 + x)}\right) $.