How do I find the derivative of this problem?

Question

y = arctan (x/a) ?

I know arctan x becomes 1/1+x^2, but somehow my textbook gets the answer to be a/(a^2+x^2) and after an hour and a half, I am giving up trying to figure this out.

Answer 1

you have to use:

    arctan f (x) = f ' (x)

...................----------------

...................1 + f (x)^2

and f (x) = x/a

Answer 2

Use the chain rule:

dy/dx = 1 / (1 + (x/a)^2) * (1/a)

= a^2 / (a^2 + x^2) * (1/a)

= a / (a^2 + x^2)

Answer 3

Yes, the derivative of y=arctanx   y'= 1/(1+x^2), but u have x/a!

The derivative of complex function is f'=u'*v'

u'=1/(1+x^2/a^2)= (a^2)/(a^2+x^2) and v'=1/a

y'= (a^2)/(x^2+a^2))* (1/a)= a/(a^2+x^2).

Answer 4

Geez that's hard. ?_?

What level of math are you in?

Answer 5

Dude your forgetting about the "a"

Your equation have to do with "x"  The one that is giving in your book.

So since its been a long time for me your answer should be the following:

(1/a)* 1/(1+(x/a)2)

(1/a)* 1/(1+(x^2/a^2))

"Now multiply the denominator and whatever they call the other thing with a^2"

(1/a)* a^2/(1+(x^2/a^2))a^2

(1/a)* a^2/(a^2+x^2)

a/(a^2+x^2)

So you have to multiply with 1/a because it affects the rate of change.  Ex: derivative of sin(3x)= 3cos(3x)  Then same old thing. But you don't plug in x in your equation you put x/a

Answer 6

Use the chain rule:

y = arctan ( x/a )

dy/dx = d/dx ( arctan ( x/a ) ) * d/dx ( x/a )

dy/dx = [ ( 1 ) / 1 + (x/a)² ] * ( 1/a )

dy/dx = [ ( 1 ) / a (1 + (x/a)² ]

dy/dx = [ ( 1 ) / a (1 + (x² / a²)) ]

dy/dx = [ ( 1 ) / (a + (ax²/a²)) ]

dy/dx = [ ( 1 ) / (a³ + ax²) / a² ]

dy/dx = [ a² / (a³ + ax²) ]

dy/dx = [ a² / a (a² + x²) ]

dy/dx = [ a / (a² + x²) ]