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

The derivative of (2x+1^{x^2-1}) is (2xln(2x+1)+(2(x^2-1))/(2x+1))(2x+1)^{x^2-1}

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

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

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

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

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

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

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

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

\int \frac{4}{1 + ( 2 x + 1 ) ^{2}} d x

\int 2 ln (3 x) d x

\int \frac{sin ( 4 e ^{- 7 x} )}{e ^{7 x}} d x

\frac{d}{d x} (\frac{( 3 x - 2 )}{5 x})

\int \frac{sin ( 2 x )}{48 + cos ^{2} ( x )} d x

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