What is the derivative of f(x)=(x^x+1)^2 ?

The derivative of f(x)=(x^x+1)^2 is 2(x^x+1)x^x(ln(x)+1)

What is the first derivative of f(x)=(x^x+1)^2 ?

The first derivative of f(x)=(x^x+1)^2 is 2(x^x+1)x^x(ln(x)+1)

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derivative of

f (x) = (x^{x} + 1)^{2}

2 (x^{x} + 1) x^{x} (ln (x) + 1)

Solution steps

\frac{d}{d x} ((x^{x} + 1)^{2})

Apply the chain rule:

2 (x^{x} + 1) \frac{d}{d x} (x^{x} + 1)

= 2 (x^{x} + 1) \frac{d}{d x} (x^{x} + 1)

\frac{d}{d x} (x^{x} + 1) = x^{x} (ln (x) + 1)

= 2 (x^{x} + 1) x^{x} (ln (x) + 1)

\frac{d}{d x} ((2 x - 1) (x^{2} - 6 x + 3))

\frac{d y}{d t} = \frac{1}{1 + y}

\int 1 + sin (2 x) d x

f (x) = x^{2}, at x = 7

y^{^{''}} + 3 y^{^{'}} + 9 y = 0

What is the derivative of f(x)=(x^x+1)^2 ?
The derivative of f(x)=(x^x+1)^2 is 2(x^x+1)x^x(ln(x)+1)
What is the first derivative of f(x)=(x^x+1)^2 ?
The first derivative of f(x)=(x^x+1)^2 is 2(x^x+1)x^x(ln(x)+1)