taking derivative...but simplification process im not aware

[ku1005]

hello

with the following i have been asked to differentiate...wich isnt a prob

y = (x^2 . tanx)/ sec x

I get my answer to

x^2 (sec^2 (x)) + 2x tan x - x^2 (tan^2 (x)) / (sec^2 (x))

the answer in the book is

(2x tan x + x^2) / sec x

....im just wondering what you do to simplify it to that???im sure its really easy and i cant see it....either that or I have done something wrong in my working...any advice appreciated!

 

[ku1005]

sorry...the denominator for my answer should only be sec x ..not squared as i have written...

 

[galactus]

Take note....\(\displaystyle x^{2}sec^{2}(x)-x^{2}tan^{2}(x)=x^{2}\)

 

[soroban]

Re: taking derivative...but simplification process im not aw

Hello, ku1005!

You may have used the Quotient Rule incorrectly . . .

Differentiate: \(\displaystyle \L\:y \:= \:\frac{x^2\cdot\tan x}{\sec x}\)

Answer: \(\displaystyle \L\:\frac{2x\tan x\,+\,x^2}{\sec x}\)

Differentiate: \(\displaystyle \L\:y' \;=\;\frac{\sec x\left[x^2\cdot\sec^2x\,+\,2x\cdot\tan x\right]\,-\,x^2\cdot\tan x\cdot\sec x\cdot\tan x}{\sec^2x}\)

. . \(\displaystyle \L y'\;=\;\frac{x^2\cdot\sec^3x\,+\,2x\cdot\sec x\cdot\tan x\,-\,x^2\cdot\sec x\cdot\tan^2x}{\sec^2x}\)

Factor: \(\displaystyle \L\:y' \;=\;\frac{x\cdot\sec x\left[x\cdot\sec^2x\,+\,2\tan x\,-\,x\cdot\tan^2x\right]}{\sec x}\)

Reduce: \(\displaystyle \L\:y' \;=\;\frac{x\left[x\left(\sec^2x\,-\,\tan^2x\right)\,+\,2\tan x\right]}{\sec x}\)

Since \(\displaystyle \sec^2x\,-\,\tan^2x\:=\:1\), we have: \(\displaystyle \L y' \;=\;\frac{x\left[x\,+\,2\tan x\right]}{\sec x}\)

Therefore: \(\displaystyle \L\:y'\;=\;\frac{2x\cdot\tan x\,+\,x^2}{\sec x}\)

 

[skeeter]

\(\displaystyle \L y' = \frac{\sec{x}(x^2 \sec^2{x} + 2x\tan{x}) - x^2 \tan{x}(\sec{x}\tan{x})}{\sec^2{x}}\)

\(\displaystyle \L y' = \frac{\sec{x}(x^2\sec^2{x} + 2x\tan{x} - x^2\tan^2{x})}{\sec^2{x}}\)

\(\displaystyle \L y' = \frac{x^2\sec^2{x} + 2x\tan{x} - x^2\tan^2{x}}{\sec{x}}\)

\(\displaystyle \L y' = \frac{x^2(\sec^2{x}-\tan^2{x}) + 2x\tan{x}}{\sec{x}}\)

now, since tan²x + 1 = sec²x, sec²x - tan²x = 1 ...

\(\displaystyle \L y' = \frac{x^2 + 2x\tan{x}}{\sec{x}}\)

btw ... things would have been alot easier if you (and the book writers) took notice that \(\displaystyle \L y = \frac{x^2 \tan{x}}{\sec{x}} = x^2 \sin{x}\)

 

[ku1005]

thanks heaps for your help....

um nah didnt use quotient rule incorrectly...i just didnt realise

"(sec^2 (x)) - (tan^2(x)) = 1"

BUT

now I do...which is great...since it helps me solve like 2 other problems in which
I was having the same trouble

also....just wondering...how do I / what progrm do I ned to write as soroban does (ie all the matehmatical symbols etc???

cheers again