What is the derivative of ln((2x+3^4)(x+1)^3) ?

The derivative of ln((2x+3^4)(x+1)^3) is 8/(2x+3)(x+1)^3+3(x+1)^2ln((2x+3)^4)

What is the first derivative of ln((2x+3^4)(x+1)^3) ?

The first derivative of ln((2x+3^4)(x+1)^3) is 8/(2x+3)(x+1)^3+3(x+1)^2ln((2x+3)^4)

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\frac{d}{d x} (ln ((2 x + 3)^{4}) (x + 1)^{3})

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

Solution steps

\frac{d}{d x} (ln ((2 x + 3)^{4}) (x + 1)^{3})

Apply the Product Rule:

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

f = ln ((2 x + 3)^{4}), g = (x + 1)^{3}

= \frac{d}{d x} (ln ((2 x + 3)^{4})) (x + 1)^{3} + \frac{d}{d x} ((x + 1)^{3}) ln ((2 x + 3)^{4})

\frac{d}{d x} (ln ((2 x + 3)^{4})) = \frac{8}{2 x + 3}

\frac{d}{d x} ((x + 1)^{3}) = 3 (x + 1)^{2}

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

d er i v a t i v e y = - 2 e^{5 x^{2}}

4 y^{^{''}} - 25 y = 0

d er i v a t i v e f (x) = 2 + x

(2 x^{3} + y^{3}) d x - 3 x y^{2} d y = 0

\frac{d}{d θ} (sin (θ) cos (θ))

What is the derivative of ln((2x+3^4)(x+1)^3) ?
The derivative of ln((2x+3^4)(x+1)^3) is 8/(2x+3)(x+1)^3+3(x+1)^2ln((2x+3)^4)
What is the first derivative of ln((2x+3^4)(x+1)^3) ?
The first derivative of ln((2x+3^4)(x+1)^3) is 8/(2x+3)(x+1)^3+3(x+1)^2ln((2x+3)^4)