- #1

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y' = tan x + x*sec^2(x)

y'' = sec^2(x) + d/dx (x*sec^2(x))

= ... + (d/dx(sec^2(x)) + sec^2(x))

= ... + (2secx + sec^2(x))

= 2sec^3 x + sec^4(x)

what did I do wrong?? in book it says y'' = something else.

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- Thread starter PrudensOptimus
- Start date

- #1

- 635

- 0

y' = tan x + x*sec^2(x)

y'' = sec^2(x) + d/dx (x*sec^2(x))

= ... + (d/dx(sec^2(x)) + sec^2(x))

= ... + (2secx + sec^2(x))

= 2sec^3 x + sec^4(x)

what did I do wrong?? in book it says y'' = something else.

- #2

mathman

Science Advisor

- 7,969

- 508

First term should have d/dx(secx)=tanx.secx multipled, i.e. first term is 2sec= ... + (2secx + sec^2(x))

- #3

HallsofIvy

Science Advisor

Homework Helper

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= ... + (d/dx(sec^2(x)) + sec^2(x))

= ... + (2secx + sec^2(x))

= 2sec^3 x + sec^4(x)

You multiplied the first "sec^2 x" instead of adding!

- #4

- 635

- 0

I found the answer this morning,

something like

2sec^2 x(tanx + 1)

something like

2sec^2 x(tanx + 1)

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