1. Consider the function y = xx (for x> 0).
a) Why does the derivative rule for xn not apply?
b) Why does the derivative rule for ax not apply?
c) What are the four magic properties of logarithms?
d) Take the ln of both sides. Use the properties of logarithms and implicit differentiation to determine y. Express your answer as a function of x only.
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a) The derivative rule for xn does not apply to the function y = xx because it is in the form of an exponential with a variable exponent, rather than a constant exponent. The rule xn states that the derivative of xn with respect to x is n*x^(n-1), but in the case of xx, the exponent depends on the value of x, making it a variable exponent.
b) The derivative rule for ax does not apply to the function y = xx because it is not in the form of a constant base raised to a variable exponent. The rule for ax states that the derivative of ax with respect to x is ln(a) * ax. However, in the function xx, both the base and the exponent are variables, and it does not have a constant base. Therefore, the derivative rule for ax does not directly apply.
c) The four magic properties of logarithms are as follows:
1. Product rule: log(a * b) = log(a) + log(b) - This property allows us to split a logarithm of a product into the sum of logarithms of the individual factors.
2. Quotient rule: log(a / b) = log(a) - log(b) - This property allows us to split a logarithm of a quotient into the difference of logarithms of the numerator and the denominator.
3. Power rule: log(a^b) = b * log(a) - This property allows us to move the exponent of the argument as a coefficient in front of the logarithm.
4. Change of base rule: logₐ(b) = log(c, b) / log(c, a) - This property allows us to change the base of a logarithm to any other base using a different logarithm.
d) Taking the natural logarithm (ln) of both sides of the equation y = xx, we have ln(y) = ln(xx). Using the property of logarithms that states ln(a^b) = b * ln(a), we can rewrite the equation as ln(y) = x * ln(x).
Now, we can apply implicit differentiation to find y'. Differentiating both sides of the equation with respect to x, we get:
1/y * dy/dx = ln(x) + 1 * (d/dx)(x * ln(x))
Simplifying further, we have:
dy/dx = y * (ln(x) + 1)
Since y = xx, we can substitute it back into the equation:
dy/dx = (xx) * (ln(x) + 1)
Therefore, the derivative y' as a function of x only is given by:
y' = x^x * (ln(x) + 1)