Question

Find dy/dx, if y = 10^tan(u) * log base 5 of (sec^2 u^3)

Answer 1

there's no x! :D

Answer 2

right, dy/du then

Answer 3

dy/dx=0

Answer 4

really?

Answer 5

hmm i am a little slow. product rule and
\[\frac{d}{du}10^{\tan(u)}=10^{\tan(u)}\sec^2(u) \ln(10)\] so far so good?

Answer 6

@imran what constant is this?

Answer 7

I am finding dy/dx , not dy/du

Answer 8

Satellite, you used the product rule?

Answer 9

@ imran lol
@soda i didn't do anything yet except take the derivative of
\[10^{\tan(u)}\]

Answer 10

you still have to take the derivative of
\[\log_5(\sec^2(u^3))\] which will require some chain rule business, but before we do that lets rewrite it as
\[2\log_5(\sec(u^3))\] to make life a tiny bit easier

Answer 11

good god this is a pain. your math teacher must hate you

Answer 12

How did you find the derivative of \[10^\tan u\] to be \[10^\tan u * \sec^2 u \ln 10\]?

Answer 13

chain rule and the fact that
\[\frac{d}{dx}10^x=10^x\ln(10)\]

Answer 14

or more generally
\[\frac{d}{dx}b^x=b^x\ln(b)\]

Answer 15

so where does the \[\sec ^2 u \] come from?

Answer 16

the ln(b) is where it comes from, right?

Answer 17

it is the derivative of
\[\tan(u)\] so by the chain rule we get
\[\frac{d}{du}10^{\tan(u)}=10^{\tan(u)}\ln(10)\times \frac{d}{du}\tan(u)\]

Answer 18

ok

Answer 19

now we still have to take the derivative of
\[2\log_5(\sec(u^3))\] again a big chain rule problem

Answer 20

would you change \[2\log_{5}(\sec u^3)\] to exponential form?

Answer 21

are you sure it is not dy/dx?

Answer 22

no

Answer 23

the derivative of
\[\log_b(x)\] is
\[\frac{1}{\ln(b)x}\] so here we will get
\[\frac{2}{\ln(5)\sec(u^3)}\times \frac{d}{du}\sec(u^3)\]
\[\frac{2\sec(u^3)\tan(u^3)\times 3u^2}{\ln(5)\sec(u^3)}\]

Answer 24

or after canceling you get
\[\frac{6u^2\tan(u^3)}{\ln(5)}\]

Answer 25

You made that easy to understand Sate, thanks!

Answer 26

and we are still not done. because you have to use the product rule for this one

\[(fg)'=f'g+g'f\] with
\[f(u)=10^{\tan(u)}, f'(u)=10^{\tan(u)}\sec^2(u) \ln(10)\]
\[g(u)=2\log_5(\sec(u^3)),g'(u)=\frac{6u^2\tan(u^3)}{\ln(5)}\]

Answer 27

i will let you do that on your own

Answer 28

Yeah, I can handle the product having the derivatives, I just had trouble trying to find the derivatives.

Answer 29

good luck
and yw