What is the derivative of y=log_{2}(x)*log_{5}(x) ?

The derivative of y=log_{2}(x)*log_{5}(x) is (2ln(x))/(ln(2)ln(5)x)

What is the first derivative of y=log_{2}(x)*log_{5}(x) ?

The first derivative of y=log_{2}(x)*log_{5}(x) is (2ln(x))/(ln(2)ln(5)x)

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derivative of

y = lo g_{2} (x) \cdot lo g_{5} (x)

\frac{2 ln ( x )}{ln ( 2 ) ln ( 5 ) x}

Solution steps

\frac{d}{d x} (lo g_{2} (x) lo g_{5} (x))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = lo g_{2} (x), g = lo g_{5} (x)

= \frac{d}{d x} (lo g_{2} (x)) lo g_{5} (x) + \frac{d}{d x} (lo g_{5} (x)) lo g_{2} (x)

\frac{d}{d x} (lo g_{2} (x)) = \frac{1}{x ln ( 2 )}

\frac{d}{d x} (lo g_{5} (x)) = \frac{1}{x ln ( 5 )}

= \frac{1}{x ln ( 2 )} lo g_{5} (x) + \frac{1}{x ln ( 5 )} lo g_{2} (x)

Simplify

\frac{1}{x ln ( 2 )} lo g_{5} (x) + \frac{1}{x ln ( 5 )} lo g_{2} (x) : \frac{2 ln ( x )}{ln ( 2 ) ln ( 5 ) x}

= \frac{2 ln ( x )}{ln ( 2 ) ln ( 5 ) x}

x \to 2 - lim (\frac{1}{x ^{2} - 4})

\frac{e ^{- 4 s}}{( s + 2 ) ^{3}}

x \to 0 + lim (\frac{2 x}{x ^{2}})

\frac{d}{d x} (\frac{5 - 4 x ^{2} + x ^{5}}{x ^{3}})

\int \frac{36}{36 + e ^{x}} d x

What is the derivative of y=log_{2}(x)*log_{5}(x) ?
The derivative of y=log_{2}(x)*log_{5}(x) is (2ln(x))/(ln(2)ln(5)x)
What is the first derivative of y=log_{2}(x)*log_{5}(x) ?
The first derivative of y=log_{2}(x)*log_{5}(x) is (2ln(x))/(ln(2)ln(5)x)