What is the derivative of f(x)=x^{2-e^x} ?

The derivative of f(x)=x^{2-e^x} is x^{2-e^x}(-e^xln(x)+(2-e^x)/x)

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

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

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derivative of

f (x) = x^{2 - e^{x}}

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

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

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

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

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

\frac{- 4}{x ^{3}}

2 x^{- 3}

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

\frac{d}{d x} (\frac{x}{( y + 1 )})

\frac{d}{d x} (a (x + 2) (x - 3)^{b})

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