What is the derivative of log_{10}(2^x+2^{-x}) ?

The derivative of log_{10}(2^x+2^{-x}) is (ln(2)(4^x-1))/(ln(10)(4^x+1))

What is the first derivative of log_{10}(2^x+2^{-x}) ?

The first derivative of log_{10}(2^x+2^{-x}) is (ln(2)(4^x-1))/(ln(10)(4^x+1))

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derivative of

lo g_{10} (2^{x} + 2^{- x})

\frac{ln ( 2 ) ( 4 ^{x} - 1 )}{ln ( 10 ) ( 4 ^{x} + 1 )}

Solution steps

\frac{d}{d x} (lo g_{10} (2^{x} + 2^{- x}))

Apply log rule:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

= \frac{d}{d x} (\frac{ln ( 2 ^{x} + 2 ^{- x} )}{ln ( 10 )})

Take the constant out:

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

= \frac{1}{ln ( 10 )} \frac{d}{d x} (ln (2^{x} + 2^{- x}))

Apply the chain rule:

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

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

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

= \frac{1}{ln ( 10 )} \cdot \frac{1}{2 ^{x} + 2 ^{- x}} (2^{x} ln (2) - ln (2) \cdot 2^{- x})

Simplify

\frac{1}{ln ( 10 )} \cdot \frac{1}{2 ^{x} + 2 ^{- x}} (2^{x} ln (2) - ln (2) \cdot 2^{- x}) : \frac{ln ( 2 ) ( 4 ^{x} - 1 )}{ln ( 10 ) ( 4 ^{x} + 1 )}

= \frac{ln ( 2 ) ( 4 ^{x} - 1 )}{ln ( 10 ) ( 4 ^{x} + 1 )}

\frac{d}{d n} (i^{2 n})

d er i v a t i v e cot (x^{4})

d er i v a t i v e x^{2} (1 - m^{2})

d er i v a t i v e f (x) = 10 x^{2} + 5 x - 2

d er i v a t i v e 2 x - y + 1

What is the derivative of log_{10}(2^x+2^{-x}) ?
The derivative of log_{10}(2^x+2^{-x}) is (ln(2)(4^x-1))/(ln(10)(4^x+1))
What is the first derivative of log_{10}(2^x+2^{-x}) ?
The first derivative of log_{10}(2^x+2^{-x}) is (ln(2)(4^x-1))/(ln(10)(4^x+1))