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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

- 2 x + e^{x - 1} - \frac{1}{x} + 2

Solution steps

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

Simplify

e^{x - 1} - ln (x) - (x - 1)^{2} - 1 : - x^{2} + 2 x + e^{x - 1} - ln (x) - 2

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d}{d x} (x^{2}) = 2 x

\frac{d}{d x} (2 x) = 2

\frac{d}{d x} (e^{x - 1}) = e^{x - 1}

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

\frac{d}{d x} (2) = 0

= - 2 x + 2 + e^{x - 1} - \frac{1}{x} - 0

Simplify

= - 2 x + e^{x - 1} - \frac{1}{x} + 2

\int_{1}^{2} 5 4 x^{2} - 3 d x

(\frac{1}{t - 1})^{^{'}}

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

y^{^{'''}} + 49 y^{^{'}} = 0

e^{x^{5}}

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