Question

if y = (sin x)^x^2 then differentiate this!!!!! Pls give full steps!!!!!!

Answer 1

chane rule

Answer 2

use d f(x)^g(x)/ dx = f(x)^g(x)d/dx(g(x)log(f(x))

This can be proved easily

Answer 3

=cos x *2x

Answer 4

so here i gotta apply the chain rule got it so i have to learn it keep answering ppl i will just go through and then maybe I will be able to understand the solution!!!!

Answer 5

keep posting what you do!

Answer 6

ok

Answer 7

@ruba: not you, I was telling that to tejeshwar so that he may understand and we may correct him if you are wrong

Answer 8

Tejeshwar: this is not exaclty chain rule, f(x)^g(x) is a bit different thing

Answer 9

nope sure its chain rule

Answer 10

First take y=sin(x)^x^2
log(y)=x^2log(sinx)

d/dx on both sides
so you get,
1/y dy/dx = d/dx (x^2log(sinx))
therefore, dy/dx=y d/dx(x^2log(sin(x))

Answer 11

\[sin(x)^{x^2}\]?

Answer 12

ohhhhhhh so so so s o sry mogh
im jst 4geting this

Answer 13

if so take the log, simplify, take the derivative, then multiply by \[sin(x)^{x^2}\]

Answer 14

ri8 ri8 sat & mogh go on nd bye

Answer 15

Take this into consideration,

y = f(x)^g(x)
log y = g(x) log(f(x))
\[(1/y)dy/dx = d/dx(g(x)\log(f(x)))\]

Answer 16

\[ln(sin(x)^{x^2})=x^2ln(sin(x))\]
\[\frac{d}{dx}x^2ln(sin(x))=2xln(sin(x))+\frac{x^2}{x}=2xln(sin(x))+x\]
answer:
\[(sin(x))^{x^2}(2xln(sin(x))+x)\]

Answer 17

oops sorry
wrong

Answer 18

\[\frac{d}{dx}x^2ln(sin(x))=2xln(sin(x))+\frac{x^2}{sin(x)}\]

Answer 19

my mistake. sorry.

Answer 20

@satellite: d/dx ln(sin(x)) is \[cosx/sinx\]

Answer 21

wow am i off . you are right and i am wrong!

Answer 22

two mistakes in one post i should resign.

Answer 23

lol, it happens!

Answer 24

btw it is also correct two write
\[(sin(x))^{x^2}=e^{x^2ln(sin(x))}\] and differentiate this using the chain rule. the actual work is identical since it all boils down to finding the derivative of
\[x^2ln(sin(x))\]CORRECTLY

Answer 25

just hang on satellite i will just be back 1ce i understand the chain rule!!!!

Answer 26

Chain rule is easier than it looks, just check out the proof too!
Regards.