- 637

- 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.

- Thread starter PrudensOptimus
- Start date

- 637

- 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.

Science Advisor

- 7,688

- 387

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

Science Advisor

Homework Helper

- 41,709

- 876

You multiplied the first "sec^2 x" instead of adding!= ... + (d/dx(sec^2(x)) + sec^2(x))

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

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

- 637

- 0

I found the answer this morning,

something like

2sec^2 x(tanx + 1)

something like

2sec^2 x(tanx + 1)

• Topics based on mainstream science

• Proper English grammar and spelling

We Value Civility

• Positive and compassionate attitudes

• Patience while debating

We Value Productivity

• Disciplined to remain on-topic

• Recognition of own weaknesses

• Solo and co-op problem solving