What is the derivative of 8/(cos(x^2-2x)) ?

The derivative of 8/(cos(x^2-2x)) is 8sec(x^2-2x)tan(x^2-2x)(2x-2)

What is the first derivative of 8/(cos(x^2-2x)) ?

The first derivative of 8/(cos(x^2-2x)) is 8sec(x^2-2x)tan(x^2-2x)(2x-2)

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\frac{d}{d x} (\frac{8}{cos ( x ^{2} - 2 x )})

8 sec (x^{2} - 2 x) tan (x^{2} - 2 x) (2 x - 2)

Solution steps

\frac{d}{d x} (\frac{8}{cos ( x ^{2} - 2 x )})

Simplify

\frac{8}{cos ( x ^{2} - 2 x )} : 8 sec (x^{2} - 2 x)

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

Take the constant out:

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

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

Apply the chain rule:

sec (x^{2} - 2 x) tan (x^{2} - 2 x) \frac{d}{d x} (x^{2} - 2 x)

= sec (x^{2} - 2 x) tan (x^{2} - 2 x) \frac{d}{d x} (x^{2} - 2 x)

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

= 8 sec (x^{2} - 2 x) tan (x^{2} - 2 x) (2 x - 2)

x cos (x) + 2 tan (x)

\int_{5}^{7} \frac{x}{x ^{2} + 6 x + 13} d x

(2 x^{2} + 2)^{3} (x^{2} - 1)^{2}

(2 x^{\frac{3}{2}})^{^{'}}

t^{2} \frac{d y}{d t} - t = 1 + y + t y, y (1) = 7

What is the derivative of 8/(cos(x^2-2x)) ?
The derivative of 8/(cos(x^2-2x)) is 8sec(x^2-2x)tan(x^2-2x)(2x-2)
What is the first derivative of 8/(cos(x^2-2x)) ?
The first derivative of 8/(cos(x^2-2x)) is 8sec(x^2-2x)tan(x^2-2x)(2x-2)