Is this correct?
2007-03-14 00:01:56 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
No it is not. Deriv of arctan hasn't any square roots.
Use (d/du)[arctan u]
= 1/(1 + u^2)
When u = 6x/5, to find dy/dx we multiply this derivative by
du/dx
Thus
dy/dx
= [1/(1 + (6x/5)^2)]*6/5
Multiply top and bottom by 25 to get
30/(25 + 36x^2)
2007-03-14 00:11:56 · answer #1 · answered by Hy 7 · 0⤊ 0⤋
The way to get the correct answer is to start by saying
y = arctan(6x/5) implies tany = 6x/5. Then use implicit differentiation giving (secy)^2*dy/dx = 6/5. Then use the identity (secy)^2 = 1 + (tany)^2 on the LHS. Take it from there.
2007-03-14 00:15:57 · answer #2 · answered by Anonymous · 0⤊ 0⤋
y = tan^(-1) (6x / 5) and let u = 6x / 5
dy/du = 1/(1 + 36x² / 25)
dy/du = 25 / (25 + 36x²)
du/dx = 6 / 5
dy/dx = (6 / 5).25 / (25 + 36x²)
dy/dx= 30 / (25 + 36.x²)
2007-03-14 00:43:04 · answer #3 · answered by Como 7 · 0⤊ 0⤋
y = acrtan 6x/5
=> tan y = 6x/5
d.w.r.t. x, we get
(sec y)^2 dy/dx = 6/5
=> dy/dx = 6/5((sec y)^2)
=> dy/dx = 6/5(1+(tany)^2)
by putting the value of tan y above we will get,
dy/dx = 6/5(1+(6x/5)^2)
=> dy/dx = 30/(25 + 36x^2)........answer
2007-03-14 00:30:49 · answer #4 · answered by desh 1 · 0⤊ 0⤋
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2016-12-01 23:43:43 · answer #5 · answered by ? 3 · 0⤊ 0⤋
I swear I knew this.. I swear..
2007-03-14 00:12:37 · answer #6 · answered by Alicia 3 · 0⤊ 0⤋