What is the derivative of f(x)=-(1/3)^{(x+2)} ?

The derivative of f(x)=-(1/3)^{(x+2)} is ln(3)(1/3)^{x+2}

What is the first derivative of f(x)=-(1/3)^{(x+2)} ?

The first derivative of f(x)=-(1/3)^{(x+2)} is ln(3)(1/3)^{x+2}

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derivative of

f (x) = - (\frac{1}{3})^{(x + 2)}

ln (3) (\frac{1}{3})^{x + 2}

Solution steps

\frac{d}{d x} (- (\frac{1}{3})^{x + 2})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= - \frac{d}{d x} ((\frac{1}{3})^{x + 2})

Apply exponent rule:

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

(\frac{1}{3})^{x + 2} = e^{(x + 2) l n (\frac{1}{3})}

= - \frac{d}{d x} (e^{(x + 2) l n (\frac{1}{3})})

Apply the chain rule:

e^{(x + 2) l n (\frac{1}{3})} \frac{d}{d x} ((x + 2) ln (\frac{1}{3}))

= e^{(x + 2) l n (\frac{1}{3})} \frac{d}{d x} ((x + 2) ln (\frac{1}{3}))

\frac{d}{d x} ((x + 2) ln (\frac{1}{3})) = - ln (3)

= - e^{(x + 2) l n (\frac{1}{3})} (- ln (3))

Simplify

- e^{(x + 2) l n (\frac{1}{3})} (- ln (3)) : ln (3) (\frac{1}{3})^{x + 2}

= ln (3) (\frac{1}{3})^{x + 2}

n \to 1 lim (\frac{1}{n ^{2}})

\int_{1}^{9} x \cdot ln (x) d x

\int x^{2} + x + 2 d x

x = 4 cos (t)

\int_{2}^{4} x \frac{( 2 x - 3 )}{6} d x

What is the derivative of f(x)=-(1/3)^{(x+2)} ?
The derivative of f(x)=-(1/3)^{(x+2)} is ln(3)(1/3)^{x+2}
What is the first derivative of f(x)=-(1/3)^{(x+2)} ?
The first derivative of f(x)=-(1/3)^{(x+2)} is ln(3)(1/3)^{x+2}