What is the derivative of 3^{xlog_{2}(x^4}) ?

The derivative of 3^{xlog_{2}(x^4}) is ln(3)*81^{xlog_{2}(x)}(4log_{2}(x)+4/(ln(2)))

What is the first derivative of 3^{xlog_{2}(x^4}) ?

The first derivative of 3^{xlog_{2}(x^4}) is ln(3)*81^{xlog_{2}(x)}(4log_{2}(x)+4/(ln(2)))

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\frac{d}{d x} (3^{x l o g_{2} (x^{4})})

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

Solution steps

\frac{d}{d x} (3^{x l o g_{2} (x^{4})})

Apply exponent rule:

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

3^{x l o g_{2} (x^{4})} = e^{x l o g_{2} (x^{4}) l n (3)}

= \frac{d}{d x} (e^{x l o g_{2} (x^{4}) l n (3)})

Apply the chain rule:

e^{x l o g_{2} (x^{4}) l n (3)} \frac{d}{d x} (x lo g_{2} (x^{4}) ln (3))

= e^{x l o g_{2} (x^{4}) l n (3)} \frac{d}{d x} (x lo g_{2} (x^{4}) ln (3))

\frac{d}{d x} (x lo g_{2} (x^{4}) ln (3)) = ln (3) (lo g_{2} (x^{4}) + \frac{4}{ln ( 2 )})

= e^{x l o g_{2} (x^{4}) l n (3)} ln (3) (lo g_{2} (x^{4}) + \frac{4}{ln ( 2 )})

Simplify

e^{x l o g_{2} (x^{4}) l n (3)} ln (3) (lo g_{2} (x^{4}) + \frac{4}{ln ( 2 )}) : ln (3) \cdot 8 1^{x l o g_{2} (x)} (4 lo g_{2} (x) + \frac{4}{ln ( 2 )})

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

\frac{d}{d x} (ln (n x + 10))

x \to 0 + lim (x^{\frac{4}{x}})

x \to 0 lim (sin (7 x))

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

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

What is the derivative of 3^{xlog_{2}(x^4}) ?
The derivative of 3^{xlog_{2}(x^4}) is ln(3)*81^{xlog_{2}(x)}(4log_{2}(x)+4/(ln(2)))
What is the first derivative of 3^{xlog_{2}(x^4}) ?
The first derivative of 3^{xlog_{2}(x^4}) is ln(3)*81^{xlog_{2}(x)}(4log_{2}(x)+4/(ln(2)))