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

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Answer 1 · 2016-09-15T05:37:43

$\frac{2}{\sqrt{4 x^{2} + 20 x + 24} \cdot i}$ $(2) / (sqrt(4 x^(2) + 20 x + 24) cdot i)$

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

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 $2 x + 5$ :

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

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

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

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

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

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