easy differentiation of trig functions q (1 Viewer)

[kloudsurfer]

Hey,
Im stuck on this question:

Find the gradient of the tangent to the curve y= tan3x at the point where x=π/9
(π is pi btw)

The answer in the book is 12, but keep getting 4/3 (which happens to be 1/9 of 12) Heres my working out:
y=tan3x
y= 3sec^2 3x
y'= 3sec^2(3π/9)
= 1/(3cos^2 (π/3))
= 4/3
therefore m=4/3

I dont know where I went wrong but im sure i just made a stupid mistake as usual.

Could someone please help me?

Thanks in advance

 

[watatank]

here's the proper solution. all you did wrong i think was put the 3 in the denominator, which isn't what you are meant to do. you seperate the constant out the front from the other stuff.

y=tan3x
y'= 3 * sec^2 3x
gradient = 3 * sec^2(3π/9), since x = π/9
= { 3 / [ cos^2 (π/3) ] }
= [ 3 / (1/2)^2 ]
= [ 3 / (1/4) ]
= 12

 

[kloudsurfer]

here's the proper solution. all you did wrong i think was put the 3 in the denominator, which isn't what you are meant to do. you seperate the constant out the front from the other stuff.

y=tan3x
y'= 3 * sec^2 3x
gradient = 3 * sec^2(3π/9), since x = π/9
= { 3 / [ cos^2 (π/3) ] }
= [ 3 / (1/2)^2 ]
= [ 3 / (1/4) ]
= 12

I knew I made a stupid mistake!

I went:
y'= 3sec^2(3π/9)
= 1/(3cos^2 (π/3))

and brought the 3 down to the bottom! I always do that lol!

Thankyou so much!!! Youre a lifesaver!