What is the derivative of (8(4^x))/(x^4) ?

The derivative of (8(4^x))/(x^4) is (16(ln(2)*4^xx-2^{2x+1}))/(x^5)

What is the first derivative of (8(4^x))/(x^4) ?

The first derivative of (8(4^x))/(x^4) is (16(ln(2)*4^xx-2^{2x+1}))/(x^5)

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derivative of

\frac{8 ( 4 ^{x} )}{x ^{4}}

\frac{16 ( ln ( 2 ) \cdot 4 ^{x} x - 2 ^{2 x + 1} )}{x ^{5}}

Solution steps

\frac{d}{d x} (\frac{8 \cdot 4 ^{x}}{x ^{4}})

Take the constant out:

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

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= 8 \cdot \frac{\frac{d}{d x} ( 4 ^{x} ) x ^{4} - \frac{d}{d x} ( x ^{4} ) 4 ^{x}}{( x ^{4} ) ^{2}}

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

\frac{d}{d x} (x^{4}) = 4 x^{3}

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

Simplify

8 \cdot \frac{ln ( 2 ) \cdot 2 ^{2 x + 1} x ^{4} - 4 x ^{3} \cdot 4 ^{x}}{( x ^{4} ) ^{2}} : \frac{16 ( ln ( 2 ) \cdot 4 ^{x} x - 2 ^{2 x + 1} )}{x ^{5}}

= \frac{16 ( ln ( 2 ) \cdot 4 ^{x} x - 2 ^{2 x + 1} )}{x ^{5}}

\int_{- 1}^{4} x - 2 x^{2} d x

\frac{\partial}{\partial x} (x y e^{y})

f (x) = 2^{s i n (x)}

y = (\frac{8}{x})^{5 x}

\frac{d}{d x} (ln (x + x^{2} + 5))

What is the derivative of (8(4^x))/(x^4) ?
The derivative of (8(4^x))/(x^4) is (16(ln(2)*4^xx-2^{2x+1}))/(x^5)
What is the first derivative of (8(4^x))/(x^4) ?
The first derivative of (8(4^x))/(x^4) is (16(ln(2)*4^xx-2^{2x+1}))/(x^5)