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1. What is the 2nd derivative of :
tan^2(2x^2 +3)

Thanks
2. (Original post by kevins)
What is the 2nd derivative of :
tan^2(2x^2 +3)

Thanks
The working is quite tedious, so I'll simply state the derivatives;

y = (tan(2x² +3))²

dy/dx = 8xsec(2x²+3)²tan(2x² +3)

d²y/dx² = -4sec(2x²+3)^4.[8cos(4x² +6)x²-16x²-sin(4x² +6)]
3. 8x.sec(2x²+3)².tan(2x²+3)
4. (Original post by manps)
8x.sec(2x²+3)².tan(2x²+3)
He wants d²y/dx²
5. (Original post by manps)
8x.sec(2x²+3)².tan(2x²+3)
2nd derivative dude.
6. [QUOTE]
Thanks but can u put it in terms on sec and tan. really appreciate it
7. [QUOTE=kevins]
Thanks but can u put it in terms on sec and tan. really appreciate it
Now you are really asking a difficult question.
8. (Original post by Dekota)
Now you are really asking a difficult question.

Some people are so lazy.
9. its jsut tha wheni do it i get my answer in terms of tan and sec and yours has sin and cos in it
10. (Original post by kevins)
its jsut tha wheni do it i get my answer in terms of tan and sec and yours has sin and cos in it
You do one thing, to check if yours is the same as mine...

Let x = 1, and substitute this into your answer for d²y/dx²

and then let x = 1, and substitute this into my answer for d²y/dx²...
d²y/dx² = -4sec(2x²+3)^4.[8cos(4x² +6)x²-16x²-sin(4x² +6)]

If both give the same answer, then obviously they are equal and the same.

11. yeh yours seems to be right def, thanks alot but would i be pushing it if i was askin for the working out. i can give you my email adress if you want.
12. (Original post by kevins)
yeh yours seems to be right def, thanks alot but would i be pushing it if i was askin for the working out. i can give you my email adress if you want.
How about, tomorrow .. I'll post it up?
13. yeh thanks mate
14. (Original post by kevins)
What is the 2nd derivative of :
tan^2(2x^2 +3)
y = [tan(2x² +3)]²
dy/dx = 2[tan(2x² +3)].d[tan(2x² +3)]/dx
.........= 2[tan(2x² +3)].4xsec(2x² +3)²
.........= 8x[tan(2x² +3)]sec(2x² +3)²

1/8.d²y/dx²....= x[tan(2x² +3)]sec(2x² +3)²
...................= x[sin(2x² +3)/cos(2x² +3)]sec(2x² +3)²
...................= xsin(2x² +3)/[cos(2x² +3)cos(2x² +3)²]
...................= d[xsin(2x² +3)]/dx.[cos(2x² +3)cos(2x² +3)²] - xsin(2x² +3)].d[cos(2x² +3)cos(2x² +3)²]/dx
...................= [x.d[sin(2x² +3)]/dx + d[x]/dx.sin(2x² +3)].[cos(2x² +3)cos(2x² +3)²] - xsin(2x² +3)].cos(2x² +3).d[cos(2x² +3)²]/dx + [d[cos(2x² +3)]/dx.cos(2x² +3)²]
...................= [4x²cos(2x² +3) + sin(2x² +3)].[cos(2x² +3)cos(2x² +3)²] - xsin(2x² +3)].[cos(2x² +3).-(16x³+24x)sin(4x^4 +12x²+6) - 4xsin(2x² +3)cos(2x² +3)²]
...................= [4x²cos(2x² +3) + sin(2x² +3)cos(2x² +3)cos(2x² +3)²] + [xsin(2x² +3)cos(2x² +3)sin(4x^4 +12x²+6) - 4xsin(2x² +3)cos(2x² +3)²]
...................= [[4x²cos(2x² +3) + sin(2x² +3)cos(2x² +3)cos(2x² +3)²] + [xsin(2x² +3)cos(2x² +3)sin(4x^4 +12x²+6) - 4xsin(2x² +3)cos(2x² +3)²]] / [cos(2x² +3)cos(2x² +3)²]²
...................= [[4x²cos(2x² +3) + sin(2x² +3)cos(2x² +3)cos(2x² +3)²] + [xsin(2x² +3)cos(2x² +3)sin(4x^4 +12x²+6) - 4xsin(2x² +3)cos(2x² +3)²]] / [cos(2x² +3)cos(2x² +3)²]²
...................= 8[[[4x²cos(2x² +3) + sin(2x² +3)cos(2x² +3)cos(2x² +3)²] + [xsin(2x² +3)cos(2x² +3)sin(4x^4 +12x²+6) - 4xsin(2x² +3)cos(2x² +3)²]] / [cos(2x² +3)cos(2x² +3)²]²]

This result is from the quotient rule. I may have made mistakes and haven't checked it yet. But after clearing this expression.. you should end up with my earlier stated expression for d²y/dx², or something similar.

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Updated: November 12, 2005
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