Question

How do you do the derivative of these functions?

Answer 1

\[y=\sin ^{2}x-tg(x ^{2}-1)\]

Answer 2

and these ones:
\[y=4\sin(\ln \frac{ x }{ 2 })\]

Answer 3

\[y=\ln(x ^{4}-3x ^{2)}\]

Answer 4

\[y=xcos ^{3}(5x+1)\]

Answer 5

\[y=\ln \frac{ 3x ^{2}-1 }{ x }\]

Answer 6

for the first one use the chain foe sin^2 x
note that tg are constants

Answer 7

number use the chain rule

Answer 8

\[y=(\sqrt{x-1})(\sin ^{2} x )\]

Answer 9

what is the chain rule?

Answer 10

nunber 2 and 3 use chain rule

Answer 11

number 4 use product rule and chain rule

Answer 12

Can you plz show me? D:

Answer 13

last one use chain and quotient rule.

Answer 14

ok chain rule or function of a function rule
example

sin 2x

2x is a function of x and sin 2x is a function of 2x Hence the above name
'
f' f(g(x)) = f'(gx) * g'(x)

so for sin 2x

derivative is cos 2x * 2 = 2 cos 2x

Answer 15

so look for a function within a function
examples would be

sin(x^2 + 8)

ln(2x^2)

log (tan x)


do you follow me?

Answer 16

ok wait

Answer 17

thank you by the way

Answer 18

another example would be cos^2 x

or ( x^2 + 8)^2

Answer 19

OK

Answer 20

So the derivative of the first function should be:
\[y=\sin ^{2}x-tg(x ^{2}-1)\]

Answer 21

y'=\[y'=2sinxcosx-(1+\tan(x ^{2}-1))2x\]

Answer 22

Is it correct? meanwhile I will do the others :)

Answer 23

2 sinx cos x is correct

Answer 24

where did you get the tan from ? does tg mean tangent?

Answer 25

yeah

Answer 26

oh ok lol - i guessed tg was just a constant

Answer 27

the derivative of tan is sec^2

so its

y' = 2 sinx cos x - 2x sec^2 ( x^2 - 1)

Answer 28

ah ok

Answer 29

As for \[y=4sinx(\ln(\frac{ x }{ 2 }))\]

Answer 30

\[y'=4cosx(\ln(\frac{ x }{ 2 }))+4sinx(\frac{ 1 }{ x })\]

Answer 31

Is this the resullt?

Answer 32

hold on - i've got confused myself

Answer 33

you have 3 functions here the sin , ln and x/2 the 4 just stays there
perform the chain rule from left to right

y' = 4 cos (ln (x/2)) * 1/ (x/2) * 1/2

Answer 34

which simplifies to
(4/x) cos (ln (x/2))

Answer 35

Umm

Answer 36

first deal with the sine , then the ln funtion and finally the 2/x - they are multiplied together

Answer 37

I applied other rules because we didn't study the chain rule in my country. I know about it but here we never use it. I applied this ones:
\[f(x)g(x)=f'(x)g(x)+f(x)g'(x)\]

Answer 38

I'll show you how I did it and thank you again xD

Answer 39

thats the product rule for the product of 2 functions but we haven't got a product here
4 sin (ln (x/2)) is not a product of functions its a compound function if u like

Answer 40

\[f(x)=4sinx\]
\[g(x)=\ln(\frac{ x }{ 2 })\]
\[f'(x)=0(sinx)+4cosx\]
\[g'(x)=\ln(\frac{ x }{ 2 })=lnx-\ln2=\frac{ 1 }{ x }-0\]
\[y'=4cosx(\ln \frac{ x }{ 2 })+4sinx(\frac{ 1 }{ x })\]

Answer 41

note : sin x ln (x/2) is a product of 2 functions but sin(ln(x/2)) is not

Answer 42

Ah ok. Sorry xD I thought it was sinx but instead it's a function of a function.

Answer 43

yes exactly - follow me now?

Answer 44

sorry maylah but i have to go now
may i ask what country you are from?

Answer 45

* naylah

Answer 46

you?

Answer 47

ok bye :D