What is the derivative of y=9sec(1/3 x) ?

The derivative of y=9sec(1/3 x) is 3sec(x/3)tan(x/3)

What is the first derivative of y=9sec(1/3 x) ?

The first derivative of y=9sec(1/3 x) is 3sec(x/3)tan(x/3)

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derivative of

y = 9 sec (\frac{1}{3} x)

3 sec (\frac{x}{3}) tan (\frac{x}{3})

Solution steps

\frac{d}{d x} (9 sec (\frac{1}{3} x))

Simplify

9 sec (\frac{1}{3} x) : 9 sec (\frac{x}{3})

= \frac{d}{d x} (9 sec (\frac{x}{3}))

Take the constant out:

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

= 9 \frac{d}{d x} (sec (\frac{x}{3}))

Apply the chain rule:

sec (\frac{x}{3}) tan (\frac{x}{3}) \frac{d}{d x} (\frac{x}{3})

= sec (\frac{x}{3}) tan (\frac{x}{3}) \frac{d}{d x} (\frac{x}{3})

\frac{d}{d x} (\frac{x}{3}) = \frac{1}{3}

= 9 sec (\frac{x}{3}) tan (\frac{x}{3}) \frac{1}{3}

Simplify

9 sec (\frac{x}{3}) tan (\frac{x}{3}) \frac{1}{3} : 3 sec (\frac{x}{3}) tan (\frac{x}{3})

= 3 sec (\frac{x}{3}) tan (\frac{x}{3})

f (x) = 2 x ln (x) + x

\int (\frac{5 - 10 x ^{2}}{4 x ^{3} - 6 x + 15}) d x

f (x) = x x^{2} + 21

x \to \infty lim (1 + 4 x)

x^{2} \frac{d y}{d x} - x y + 1 = 0

What is the derivative of y=9sec(1/3 x) ?
The derivative of y=9sec(1/3 x) is 3sec(x/3)tan(x/3)
What is the first derivative of y=9sec(1/3 x) ?
The first derivative of y=9sec(1/3 x) is 3sec(x/3)tan(x/3)