What is the ((e^{2x}-5x)^2)^' ?

The ((e^{2x}-5x)^2)^' is 2(e^{2x}-5x)(e^{2x}*2-5)

What is the first ((e^{2x}-5x)^2)^' ?

The first ((e^{2x}-5x)^2)^' is 2(e^{2x}-5x)(e^{2x}*2-5)

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((e^{2 x} - 5 x)^{2})^{^{'}}

2 (e^{2 x} - 5 x) (e^{2 x} \cdot 2 - 5)

Solution steps

((e^{2 x} - 5 x)^{2})^{'}

Apply the chain rule:

2 (e^{2 x} - 5 x) (e^{2 x} - 5 x)^{'}

= 2 (e^{2 x} - 5 x) (e^{2 x} - 5 x)^{'}

(e^{2 x} - 5 x)^{'} = e^{2 x} \cdot 2 - 5

= 2 (e^{2 x} - 5 x) (e^{2 x} \cdot 2 - 5)

(1, 1), (3, - \frac{5}{6})

\int sin (3 x) sin (8 x) d x

f (x) = 3 x + 4, (4, 4)

\int sin (4 x + 1) d x

\frac{\partial}{\partial y} (sin (y) sin (x))

What is the ((e^{2x}-5x)^2)^' ?
The ((e^{2x}-5x)^2)^' is 2(e^{2x}-5x)(e^{2x}*2-5)
What is the first ((e^{2x}-5x)^2)^' ?
The first ((e^{2x}-5x)^2)^' is 2(e^{2x}-5x)(e^{2x}*2-5)