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

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

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

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

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derivative of

y = (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)

\int 3 x^{2} (x^{3} + 5)^{7} d x

\int \frac{x - 5}{( x + 4 ) ( x + 3 ) ^{2}} d x

\int_{0}^{1} (3 x + 1)^{5} d x

x^{2} cos (2 x)

e^{x} (- 8 - 3 x + 2 x^{2})

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