What is the derivative of y=x^{2x+1} ?

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

What is the first derivative of y=x^{2x+1} ?

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

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derivative of

y = x^{2 x + 1}

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

Solution steps

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

Apply exponent rule:

a^{b} = e^{b l n (a)}

x^{2 x + 1} = e^{(2 x + 1) l n (x)}

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

Apply the chain rule:

e^{(2 x + 1) l n (x)} \frac{d}{d x} ((2 x + 1) ln (x))

= e^{(2 x + 1) l n (x)} \frac{d}{d x} ((2 x + 1) ln (x))

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

= e^{(2 x + 1) l n (x)} (2 ln (x) + \frac{2 x + 1}{x})

e^{(2 x + 1) l n (x)} = x^{2 x + 1}

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

\int - \frac{5}{x} d x

\int - 6 x^{- 1} d x

y^{^{''}} + y^{^{'}} = 6 x

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

f (x) = \frac{x}{1 - ln ( x - 2 )}

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