Differentiation help please :)

'find the gradient of the curve y=sec3x at the point where x=π/9' (pi over 9)

I differentiated and got 2sec2xtan2x and then putting x in...
2 sec2(π/9)tan2(π/9) and it's not working, I'd also prefer working it out without a calculator, any help and tips please?

Also another Q i'm struggling with is
If y= 2x+3/x^1/2 what is dy/dx ?
I've differentiated and got 2x-3/2x^3/2 which is wrong :/

Steps and methods would be much appreciated with these 2 Qs, many many thanks

By sec3x do you mean $\sec^3 x$ or $\sec 3x$?

the 2nd one

Do you know the identity $\frac{dy}{dx}=\frac{dy}{du} \times \frac{du}{dx}$?

If so, make the substitution $u=3x$. It should become a lot clearer from there.

isn't it differentiated 2sec2xtan2x ?

isn't it differentiated 2sec2xtan2x ?

well if you do y=sec3x

and let y=secU, U=3x

dy/dU=secUtanU and dU/dx=3

therefore, dy/dx=3xsecUtanU+3secU

then use the subsititution: y=3xsec3xtan3x+3sec3x

What does this mean? $\frac {d}{dx} \sec 3x$ looks nothing like that

My bad: editted

For a minute there I thought I had woefully misread the question and that I was being utterly unhelpful

well if you do y=sec3x

and let y=secU, U=3x

dy/dU=secUtanU and dU/dx=3

therefore, dy/dx=3secUtanU

then use the subsititution: dy/dx= 3sex3xtan3x

Do you know the identity $\frac{dy}{dx}=\frac{dy}{du} \times \frac{du}{dx}$?

If so, make the substitution $u=3x$. It should become a lot clearer from there.

doh! I meant a 3 where I put the 2s!!!!! -_- I think that's the same answer I got Many thanks guys, v. much appreciated