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Differentiation question (trig)

SunDun111

Would the Dy/DX of Sec squared 3x be 3 Sec sqaured 3x 3Tan Squared 3x

Reply 1

jjsnyder

Original post by SunDun111

Would the Dy/DX of Sec squared 3x be 3 Sec sqaured 3x 3Tan Squared 3x

It would be 6tan(3x)sec^2(3x)

(edited 8 years ago)

Reply 2

Zacken

Original post by SunDun111

Would the Dy/DX of Sec squared 3x be 3 Sec sqaured 3x 3Tan Squared 3x

No, why do you think that?

\displaystyle \frac{d}{dx} (\sec 3x)^2 = 2 \times (\sec 3x)^{2-1} \times \frac{d}{dx}(\sec 3x)

and

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

Reply 3

Zacken

Original post by jjsnyder

It would be 2(3secxtanx)^2

Are you sure?

Reply 4

SunDun111

Original post by Zacken

No, why do you think that?

\displaystyle \frac{d}{dx} (\sec 3x)^2 = 2 \times (\sec 3x)^{2-1} \times \frac{d}{dx}(\sec 3x)

and

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

Because my teacher didnt teach us the DY/DX of Cot, Cosec and Sec so I am going over it. I thought it would be a 3 Tan 3x but I get it now thank you!

Reply 5

jjsnyder

Original post by Zacken

Are you sure?

Aha no totally misread the question :P

Reply 6

Zacken

Original post by SunDun111

Because my teacher didnt teach us the DY/DX of Cot, Cosec and Sec so I am going over it. I thought it would be a 3 Tan 3x but I get it now thank you!

Just rewrite

\sec 3x = \left(\cos 3x\right)^{-1}

and then use the chain rule. Same applies for the other two reciprocal functions.

Reply 7

DylanJ42

Original post by SunDun111

Because my teacher didnt teach us the DY/DX of Cot, Cosec and Sec so I am going over it. I thought it would be a 3 Tan 3x but I get it now thank you!

they are just reciprocal of tan, sin and cos so just differentiate them in the tan/sin/cos form.

Edit: whooops

Reply 8

SunDun111

Original post by Zacken

Just rewrite

\sec 3x = \left(\cos 3x\right)^{-1}

and then use the chain rule. Same applies for the other two reciprocal functions.

Quick question, once I differentiate a trig function, it asks me to show it is an increasing function? What do I do.

Reply 9

Zacken

Original post by SunDun111

Quick question, once I differentiate a trig function, it asks me to show it is an increasing function? What do I do.

A function

f(x)

is increasing if its derivative

f'(x) > 0

for all

x

.

So can you show that your derivative is always positive? Maybe it's a squared quantity or something.

Reply 10

SunDun111

Original post by Zacken

A function

f(x)

is increasing if its derivative

f'(x) > 0

for all

x

.

So can you show that your derivative is always positive? Maybe it's a squared quantity or something.

Well its 2Sec2x Tan2x + 2Sec Squared 2x

Reply 11

Zacken

Original post by SunDun111

Well its 2Sec2x Tan2x + 2Sec Squared 2x

What's the original function?

Reply 12

SunDun111

Original post by Zacken

What's the original function?

The original function was Sec 2x + Tan 2x I had to Differentiate it, it was defined by -pi/4 < x < pi/4

Reply 13

Zacken

Original post by SunDun111

The original function was Sec 2x + Tan 2x I had to Differentiate it, it was defined by -pi/4 < x < pi/4

Okay, so that got you

2\sec 2x \tan 2x +2\sec^2 2x

as your rightly said. Can you show that that is always positive (for the given interval of x, obviously).

(edited 8 years ago)

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Differentiation question (trig)

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