- #1

- 565

- 2

How do I get the derivative of

y=tan4x

The answer in the back of the text book

says (4/cos^2(4X))

y=tan4x

The answer in the back of the text book

says (4/cos^2(4X))

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- Thread starter Ry122
- Start date

- #1

- 565

- 2

How do I get the derivative of

y=tan4x

The answer in the back of the text book

says (4/cos^2(4X))

y=tan4x

The answer in the back of the text book

says (4/cos^2(4X))

- #2

mjsd

Homework Helper

- 726

- 3

remember tan x can be written in terms of sin and cos

- #3

- 565

- 2

yes i know that tanx= sinx/cosx

y=tan4x -> y=sin4x/cos4x

Then what?

y=tan4x -> y=sin4x/cos4x

Then what?

- #4

Gib Z

Homework Helper

- 3,346

- 6

Also use the chain rule, u=4x.

- #5

- 565

- 2

u=4x

u'=4

y=sinu/cosu

y=cosu/-sinu

y=4cosu/-4sinu

y=4cos4x/-4sin4x

What am I doing wrong?

- #6

Gib Z

Homework Helper

- 3,346

- 6

If you have a function f(x)=u/v then the derivative is f'(x)=[tex]\frac{u'v-v'u}{v^2}[/tex]. So let u=sin x and v = cos x for this, you can find the derivative of tan x is (sec x)^2, then after that use the chain rule.

- #7

- 565

- 2

u=sin4x

u'=4cos4x

v=cos4x

v'=-4sin4x

(-4sinX)(cos4X)--(4sinX)(sin4X)/(cos4X)^2

- #8

Gib Z

Homework Helper

- 3,346

- 6

One you know that, THEN use the chain rule. Do you know what the chain rule is?

- #9

- 670

- 3

Perhaps the 6th video would be useful for you to watch.

- #10

- 565

- 2

Do you mind showing me the whole working out yourself?

That would help me alot.

Thanks

That would help me alot.

Thanks

- #11

- 59

- 0

If you still are unable to complete the problem, seek help from your teacher, peers, or get a tutor.

- #12

- 182

- 0

[tex]y=tan4x ; u=4x[/tex]

[tex]\frac{dy}{dx}=\frac{d(tanu)}{du}.\frac{du}{dx}[/tex]

[tex]\Rightarrow \ \frac{dy}{dx}= sec^2u. 4 [/tex]

[tex]\Rightarrow \ \frac{dy}{dx}=4sec^{2}4x[/tex]

[tex]\frac{dy}{dx}=\frac{d(tanu)}{du}.\frac{du}{dx}[/tex]

[tex]\Rightarrow \ \frac{dy}{dx}= sec^2u. 4 [/tex]

[tex]\Rightarrow \ \frac{dy}{dx}=4sec^{2}4x[/tex]

Last edited:

- #13

- 565

- 2

The answer i need to get is in this form

(4/cos^2(4X))

I've finally figured out how to do this i just don't understand how

(4cos4x^2) + (4sin4x^2) / (4cos4x^2) = (4/cos^2(4X))

What happens to the cos and sin squared in the numerator?

- #14

hotvette

Homework Helper

- 996

- 5

Hint: isn't there a trig identity relating sin^{2} and cos^{2}?

- #15

- 565

- 2

too easy

thanks guys

thanks guys

- #16

- 182

- 0

i thought you would know [itex] \ sec^2x=\frac{1}{cos^2x} [/itex] :uhh:

The answer i need to get is in this form

(4/cos^2(4X))

I've finally figured out how to do this i just don't understand how

(4cos4x^2) + (4sin4x^2) / (4cos4x^2) = (4/cos^2(4X))

What happens to the cos and sin squared in the numerator?

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