What is the derivative of f(x)=ln(4^x+8^x) ?

The derivative of f(x)=ln(4^x+8^x) is (ln(2)*2^{2x+1}+3ln(2)*8^x)/(4^x+8^x)

What is the first derivative of f(x)=ln(4^x+8^x) ?

The first derivative of f(x)=ln(4^x+8^x) is (ln(2)*2^{2x+1}+3ln(2)*8^x)/(4^x+8^x)

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derivative of

f (x) = ln (4^{x} + 8^{x})

\frac{ln ( 2 ) \cdot 2 ^{2 x + 1} + 3 ln ( 2 ) \cdot 8 ^{x}}{4 ^{x} + 8 ^{x}}

Solution steps

\frac{d}{d x} (ln (4^{x} + 8^{x}))

Apply the chain rule:

\frac{1}{4 ^{x} + 8 ^{x}} \frac{d}{d x} (4^{x} + 8^{x})

= \frac{1}{4 ^{x} + 8 ^{x}} \frac{d}{d x} (4^{x} + 8^{x})

\frac{d}{d x} (4^{x} + 8^{x}) = ln (2) \cdot 2^{2 x + 1} + 3 ln (2) \cdot 8^{x}

= \frac{1}{4 ^{x} + 8 ^{x}} (ln (2) \cdot 2^{2 x + 1} + 3 ln (2) \cdot 8^{x})

Simplify

\frac{1}{4 ^{x} + 8 ^{x}} (ln (2) \cdot 2^{2 x + 1} + 3 ln (2) \cdot 8^{x}) : \frac{ln ( 2 ) \cdot 2 ^{2 x + 1} + 3 ln ( 2 ) \cdot 8 ^{x}}{4 ^{x} + 8 ^{x}}

= \frac{ln ( 2 ) \cdot 2 ^{2 x + 1} + 3 ln ( 2 ) \cdot 8 ^{x}}{4 ^{x} + 8 ^{x}}

What is the derivative of f(x)=ln(4^x+8^x) ?
The derivative of f(x)=ln(4^x+8^x) is (ln(2)*2^{2x+1}+3ln(2)*8^x)/(4^x+8^x)
What is the first derivative of f(x)=ln(4^x+8^x) ?
The first derivative of f(x)=ln(4^x+8^x) is (ln(2)*2^{2x+1}+3ln(2)*8^x)/(4^x+8^x)