Question

find derivative of y=tan8x + cos1/8x

Answer 1

derivative of tanx = sec^2 x
derivative the cosx = -sinx

Answer 2

is this
\[\tan(8x)+\cos(\frac{1}{8x})\]

Answer 3

so do i use the product rule

Answer 4

then you do the chain rule
8sec^(8x)-1/8sin(1/8x)

Answer 5

if so use the chain rule. the derivative of
\[\tan(x)\] is \[\sec^2(x)\] so derivative of
\[\tan(8x)=8\sec^2(8x)\]

Answer 6

8sec^2(8x)-1/8sin(1/8x)

Answer 7

@anilorap should be should be + not -

Answer 8

derivative of
\[\frac{1}{8x}\] is \[-\frac{1}{8x^2}\]

Answer 9

Q. is the x in the denominator?

Answer 10

so first term of your derivative is
\[8\sec^2(8x)\] and second term is
\[\frac{1}{8x^2}\times \sin(\frac{1}{8x})\]

Answer 11

oh good point!

Answer 12

sorry i was assuming it was what i wrote. but if not then i am wrong and anilorap is right!

Answer 13

X is next to 1/8

Answer 14

\[\cos(\frac{x}{8})\] or
\[\cos(\frac{1}{8x})\]?

Answer 15

ok

Answer 16

then i am wrong and anilorap is correct.

Answer 17

1\[\div\]8X

Answer 18

(1/8)x

Answer 19

forget that last one typed wrong

Answer 20

well in that case my answer is correct

Answer 21

ayayay.. is the xx is in the denominator yes,, he is lol

Answer 22

did u get it?