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

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

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

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

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derivative of

y = (x - 2)^{x + 1}

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

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^{(x + 1) l n (x - 2)}

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

Apply the chain rule:

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

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

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

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

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

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

\frac{d x}{d t} = x^{2} - x

\int \frac{1}{x ^{2} + 64} d x

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

f (x) = \frac{x ^{2} + 2 x - 5}{( x + 1 ) ^{2}}

t = \frac{1 - y ^{2}}{y ^{^{'}}}

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