How do you find the derivative of #f(x) = arcsin (2x + 5)#?

Question

2578 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-15T05:37:43

#(2) / (sqrt(4 x^(2) + 20 x + 24) cdot i)#

We have: #f(x) = arcsin(2 x + 5)#

This function can be differentiated using the "chain rule".

Let #u = 2 x + 5 => u' = 2# and #v = arcsin(u) => v' = (1) / (sqrt(1 - u^(2)))#:

#=> (d) / (dx) (arcsin(2 x + 5)) = 2 cdot (1) / (sqrt(1 - u^(2)))#

#=> (d) / (dx) (arcsin(2 x + 5)) = (2) / (sqrt(1 - u^(2)))#

We can now replace #u# with #2 x + 5#:

#=> (d) / (dx) (arcsin(2 x + 5)) = (2) / (sqrt(1 - (2 x + 5)^(2)))#

#=> (d) / (dx) (arcsin(2 x + 5)) = (2) / (sqrt(1 - (4 x^(2) + 20 x + 25))#

#=> (d) / (dx) (arcsin(2 x + 5)) = (2) / (sqrt(- 4 x^(2) - 20 x - 24))#

#=> (d) / (dx) (arcsin(2 x + 5)) = (2) / (sqrt(- (4 x^(2) + 20 x + 24)))#

#=> (d) / (dx) (arcsin(2 x + 5)) = (2) / (sqrt(4 x^(2) + 20 x + 24) cdot i)#