Differentiating Trigonometric Functions: Find the Derivatives

[nephi37]

find the derivatives

of differentiation of trigonometric functions

1. y=cos(3x^2+8x-2)

2. y=tan^3 2x

3. y=sin5x sin^5 x

4. y=Square root of 4sin^2x+9cos^2x

help here please..

i can't understand trigonometric functions

sorry admin or moderator, i just search the net on how to do this problems.. and this are my answer

for
1. -6x+8sin(3x^2+8x-2)

2. 6 tan 2x sec^2 2x

3. 5 cos 5x 5 sin x cos x or 5 cos 5x 5 sin^4 x cos x

4. 8 sec^4 2x tan 2x

Are my answers correct.. Help me please

 

[grzz]

let us do it one by one:
(1) is not correct.

 

[nyr91188]

(2) is basically correct except that it should be tan^2 since the derivative of tan^3 is 3tan^2...

 

[grzz]

1. y=cos(3x^2+8x-2)

re 1.

1st step: derivative of cos(...)
2nd step: MULTIPLY by the derivative of (3x^2+8x-2)

 

[SammyS]

Actually, (1) is nearly correct.

Use parentheses around (6x+8), the derivative of 3x^2+8x-2 .

Start (3) by using the product rule.


By the Way: Until you get the hang of this, it's probably best to write trig functions raised to a power as:

(sin(x))⁵ rather than sin⁵(x),

(tan(2x))³ rather than tan³(2x)

 

[nephi37]

so my answer in number 1 should be

-(6x+8)sin(3x^2+8x-2)

and in number to should be

6 tan^2 2x sec^2 2x

Am i correct.. or still wrong :)

 

[nephi37]

in number 1

-sin(3x^2+8x-2)(6x+8) then multiply -sin to (6x+8)

-(6x+8)sin(3x^2+8x-2)

am i correct now guys :)

 

[SammyS]

in number 1

-sin(3x^2+8x-2)(6x+8) then multiply -sin to (6x+8)

-(6x+8)sin(3x^2+8x-2)

am i correct now guys :)

Either one is OK !

 

[nephi37]

thanks sammy..

its now 9 am here we nid to pass this by 10am

thanks for help...
but how about my number 3 and number 4

is that correct? I am not sure what I've done..

especially number 4 with square root thing :)

 

[SammyS]

Look at √x as x^1/2,

So y=√(4sin^2x+9cos^2x)

becomes y=(4sin^2x+9cos^2x)^1/2.

Have fun.

 

[SammyS]

For #3. Use the product rule.

y = {sin5x}×{(sin x)⁵}