# Is this Correct

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

i cant 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

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

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

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
Staff Emeritus
Homework Helper
Gold Member
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)

Last edited:
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 :)

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
Staff Emeritus
Homework Helper
Gold Member
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 !

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? im not sure what i've done..

especially number 4 with square root thing :)

SammyS
Staff Emeritus
Homework Helper
Gold Member
Look at √x as x1/2,

So y=√(4sin^2x+9cos^2x)

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

Have fun.

SammyS
Staff Emeritus