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

The derivative of (x+2^{x^2-2x+1}) is (x^2-2x+1+2x^2ln(x+2)+2xln(x+2)-4ln(x+2))(x+2)^{x^2-2x}

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

The first derivative of (x+2^{x^2-2x+1}) is (x^2-2x+1+2x^2ln(x+2)+2xln(x+2)-4ln(x+2))(x+2)^{x^2-2x}

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

(x^{2} - 2 x + 1 + 2 x^{2} ln (x + 2) + 2 x ln (x + 2) - 4 ln (x + 2)) (x + 2)^{x^{2} - 2 x}

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

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

Simplify

e^{(x^{2} - 2 x + 1) l n (x + 2)} ((2 x - 2) ln (x + 2) + \frac{1}{x + 2} (x^{2} - 2 x + 1)) : (x^{2} - 2 x + 1 + 2 x^{2} ln (x + 2) + 2 x ln (x + 2) - 4 ln (x + 2)) (x + 2)^{x^{2} - 2 x}

= (x^{2} - 2 x + 1 + 2 x^{2} ln (x + 2) + 2 x ln (x + 2) - 4 ln (x + 2)) (x + 2)^{x^{2} - 2 x}

x \to - 7 - lim (\frac{x + 3}{x + 7})

\int \frac{t ^{2} + 1}{t} d t

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

\frac{d}{d x} (\frac{cos ( x )}{sec ( x )})

\int \frac{e ^{2 x}}{1 - e ^{x}} d x

What is the derivative of (x+2^{x^2-2x+1}) ?
The derivative of (x+2^{x^2-2x+1}) is (x^2-2x+1+2x^2ln(x+2)+2xln(x+2)-4ln(x+2))(x+2)^{x^2-2x}
What is the first derivative of (x+2^{x^2-2x+1}) ?
The first derivative of (x+2^{x^2-2x+1}) is (x^2-2x+1+2x^2ln(x+2)+2xln(x+2)-4ln(x+2))(x+2)^{x^2-2x}