Question

Please help me in solving this question.

find f'(x) if f(x) = tan(3x+1).

Answer 1

I have tried this question

Answer 2

Using the first principle of differentiation

Answer 3

\[\frac{d}{dx}f(g(x))=g \prime(x)f \prime(g(x))\]

Answer 4

f'(x) = lim h---> 0 f(x+h) -f(h)/h

= lim h---> 0 tan(3x+1+h) - tan(3x+1)/h

= lim h---->0 {sin (3x+1+h)/cos(3x+1+h) - sin (3x+1)/cos(3x+1)}/h

Answer 5

@mukushla we have to use first principle for solving this

Answer 6

ok

Answer 7

@vishweshshrimali5 you have expanded f(x+h) wrongly..

Answer 8

= lim h------> 0 {sin(3x+1+h)cos(3x+1) - sin(3x+1)cos(3x+1+h)}/h(cos(3x+1)cos(3x+1+h))

Answer 9

how

Answer 10

@shubhamsrg please tell me where am I mistaken

Answer 11

@shubhamsrg I got that

Answer 12

hmm.

Answer 13

thanks a lot

it should be f(x+h)=tan (3(x+h)+1)

Answer 14

right ??

Answer 15

yep..the very same

Answer 16

thanks a ton

Answer 17

have you considered the sum of 2 angles expansion

tan (3(x + h) + 1) = tan ((3x + 1) + 3h) = [tan (3x +1) + tan (3h)]/[(1 - tan(3x +1)tan(3h)]

Answer 18

f(h)=tan(3h+1)

Answer 19

It's horrible here :|
It's NOT
f'(x) = lim h---> 0 [f(x+h) -f(h)]/h

But it is
f'(x) = lim h---> 0 [f(x+h) -f(x)]/h

and f(x+h) is NOT
tan(3x+1+h)

But it is tan[3(x+h)+1].

Answer 20

This should be correct...

Answer 21

Thanks everyone

I solved it after that