Question

f(x)=(2x+3)^−2, f'(x)=?

Answer 1

do you know chain rule ?

Answer 2

i got -2(2x+3^1*(2)

Answer 3

-2(2x+3)^-1*2

Answer 4

yes, I have learnt chain rule just before.

Answer 5

how did the exponent become -1 ??
-2-1 =... ?

Answer 6

almost right, but the exponent rules says that n-1, in this case n=-2, so then -2-1=-3

Answer 7

AHHH I see!!

Answer 8

everything else is correct...

Answer 9

Thank you.
w(r)=sqrt(r^a+1)
I have got 1/2(r^a+1)^(-1/2) where did it miastake?

Answer 10

sorry i mean a is 9

Answer 11

chain rule says to diff. inner function also, so here, derivative of r^a +1 will get multiplied.

Answer 12

1/ [2 sqrt(r^9+1)] [d/dx (r^9+1)]

Answer 13

(1/2) (r^9+1)^(-1/2) * (.....)

Answer 14

9r?

Answer 15

for x^n its n x^{n-1}
for r^9, it'll be .... ?
here, x=r, n=9

Answer 16

9r^8?

Answer 17

thats correct :)

Answer 18

(1/2) (r^9+1)^(-1/2) * (9r^8)

Answer 19

thank you,
one more question how do you do
f(x)=x^(-2)sec(2/x)?
i got -2x^-3(secxtanx2/x)

Answer 20

you'll need product rule here...

Answer 21

also you forgot to differentiate 2/x ....