What is the derivative of (4x+3^4(x+1)^{-3}) ?

The derivative of (4x+3^4(x+1)^{-3}) is (16(4x+3)^3)/((x+1)^3)-(3(4x+3)^4)/((x+1)^4)

What is the first derivative of (4x+3^4(x+1)^{-3}) ?

The first derivative of (4x+3^4(x+1)^{-3}) is (16(4x+3)^3)/((x+1)^3)-(3(4x+3)^4)/((x+1)^4)

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\frac{d}{d x} ((4 x + 3)^{4} (x + 1)^{- 3})

\frac{16 ( 4 x + 3 ) ^{3}}{( x + 1 ) ^{3}} - \frac{3 ( 4 x + 3 ) ^{4}}{( x + 1 ) ^{4}}

Solution steps

\frac{d}{d x} ((4 x + 3)^{4} (x + 1)^{- 3})

Apply the Product Rule:

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

f = (4 x + 3)^{4}, g = (x + 1)^{- 3}

= \frac{d}{d x} ((4 x + 3)^{4}) (x + 1)^{- 3} + \frac{d}{d x} ((x + 1)^{- 3}) (4 x + 3)^{4}

\frac{d}{d x} ((4 x + 3)^{4}) = 16 (4 x + 3)^{3}

\frac{d}{d x} ((x + 1)^{- 3}) = - \frac{3}{( x + 1 ) ^{4}}

= 16 (4 x + 3)^{3} (x + 1)^{- 3} + (- \frac{3}{( x + 1 ) ^{4}}) (4 x + 3)^{4}

Simplify

16 (4 x + 3)^{3} (x + 1)^{- 3} + (- \frac{3}{( x + 1 ) ^{4}}) (4 x + 3)^{4} : \frac{16 ( 4 x + 3 ) ^{3}}{( x + 1 ) ^{3}} - \frac{3 ( 4 x + 3 ) ^{4}}{( x + 1 ) ^{4}}

= \frac{16 ( 4 x + 3 ) ^{3}}{( x + 1 ) ^{3}} - \frac{3 ( 4 x + 3 ) ^{4}}{( x + 1 ) ^{4}}

\frac{d q}{d t} + \frac{1}{5 \cdot 1 0 ^{- 9}} q = \frac{200}{1000}

\int \frac{( ln ^{6} ( x ) )}{x} d x

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

f (x) = (1 + 5 x)^{9}, (0, 1)

e^{- x}, 1

What is the derivative of (4x+3^4(x+1)^{-3}) ?
The derivative of (4x+3^4(x+1)^{-3}) is (16(4x+3)^3)/((x+1)^3)-(3(4x+3)^4)/((x+1)^4)
What is the first derivative of (4x+3^4(x+1)^{-3}) ?
The first derivative of (4x+3^4(x+1)^{-3}) is (16(4x+3)^3)/((x+1)^3)-(3(4x+3)^4)/((x+1)^4)