f(x)= x^(x+1) derivative please?

Question

please show steps for 10 points thanks

Answer 1

For simplicity purposes, I'm going to let f(x) = y.

y = x^(x + 1)

This is a function to a function, so we must use logarithmic differentiation. Take the ln of both sides,

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

Use the log property that allows us to bring the (x + 1) down outside of the log.

ln(y) = (x + 1)ln(x)

Differentiate implicitly. Remember to use the product rule.

(1/y)(dy/dx) = (1)ln(x) + (x + 1)(1/x)

(1/y)(dy/dx) = ln(x) + 1 + (1/x)

Multiply both sides by y,

dy/dx = y [ ln(x) + 1 + (1/x)]

But y = x^(x + 1), so

dy/dx = x^(x + 1) [ ln(x) + 1 + (1/x) ]

Side note: You're going to have three types of functions involving exponents:

1) f(x) = x^n, where n is a constant. In this case, you would use the power rule.

2) f(x) = a^x, where a is a constant. In this case, you would use the exponential derivatives; f'(x) = a^x ln(a).

3) f(x) = [g(x)]^[h(x)], which is a function to the power of a function. This is where you would use logarithmic differentiation (as done above).

Answer 2

The first answer is wrong, since the rule cannot be applied for a variable exponent.

Let y= x ^ (x+1)
ln y = (x+1) ln x

Differentiate both sides with respect to x (use product rule on the RHS):

(1/y) (dy/dx) = (x+1) (1/x) + (1)(ln x)

Which gives:

dy/dx= y { (x+1)/x + ln x}
= { x^(x+1) } { (x+1)/x + ln x}

Answer 3

If f(x) = x^(x+1) then f(x) = (x^x)(x^1), or f(x) = x*(x^x).

Therefore, by the multiplication rule of derivatives:
f '(x) = 1*(x^x) + x*((d/dx)x^x)

Let y = x^x.
Therefore ln y = ln x^x.
ln y = x*ln x
Using implicit differentiation:
(1/y)dy = ((1*ln x) + x*(1/x))dx
(1/y)dy = (ln x + 1)dx
dy/dx = y(1 + ln x)
dy/dx = (x^x)(1+ln x)

So f '(x) = 1*(x^x) + x*(x^x)(1+ln x)
f '(x) = (x^x)(1 + x(1+ln x))
f '(x) = (x^x)(1 + x + xln x)

NOTE: You cannot use the power rule ((d/dx)x^n) = n(x^(n-1)) since the exponent (x + 1) is not constant.

Answer 4

f(x)= x^(x+1)
ok. the derivitive is :
f(x)= a*n *x^{n-1}
f(x)= x+1*1* x ^ x
f(x)= x+1*x^x
Derivitive is (x+1)x^x
P.S: The representation used is not the actual formula but a representation.

Answer 5

f(x) = x^(x+1)

From x^y = e^(log(x)*y)

f(x) = e^(log(x)*(x+1))

From derivative of e^y is (e^y)*y'
And derivative of u*v is u'*v + u*v'
And derivative of log(x) is 1/x
And derivative of (x+1) is 1

f'(x) = (e^(log(x)*(x+1)))*((x+1)/x + log(x))

f'(x) = x^(x+1) * (1 + 1/x + log(x))

f'(x) = x^(x+1)*(1 + log(x)) + x^x

Answer 6

f(x) = x ^ (x +1)
f'(x) = (x+1)*x^(x+1-1) *Dx(x+1)
f'(x) = (x+1)*x^x*(1+0)
f'(x) = (x+1) * x^x