Since composing inverse functions gets us back where we started, sin(arcsin x) = x. Taking the derivative of each side of this equation gives us cos(arcsin x) · (arcsin x)' = 1 therefore ![](https://media1.shmoop.com/images/calculus/cal_compderiv_morecompl_narr_latek_34.png)
This is correct, but we're not done. cos(arcsin x) is messy, and we can simplify it. On the right-hand side, we have (x)' = 1. On the left-hand side, we have an outside function sin(□) and an inside function arcsin x. We need to use the chain rule. The derivative of sin(□) is cos(□), therefore (sin(arcsin x))' = cos(arcsin x) · (arcsin x)'. The simplification is the part that relies on SOHCAHTOA and the Pythagorean theorem. arcsin x is an angle. Call that angle θ. Then θ = arcsin x means sin θ = x. Since , we know ![](https://media1.shmoop.com/images/calculus/cal_compderiv_morecompl_narr_latek_33.png) A triangle representing this equation would look something like this: ![](https://media1.shmoop.com/images/calculus/cal_compderiv_morecompl_narr_graphik_1.png)
Using the Pythagorean Theorem we can find the missing side of the triangle (in terms of x): ![](https://media1.shmoop.com/images/calculus/cal_compderiv_morecompl_narr_graphik_2.png)
We want to know cos(arcsin x) so we can simplify the equation ![](https://media1.shmoop.com/images/calculus/cal_compderiv_morecompl_narr_latek_34.png)
If arcsin x = θ, then cos (arcsin x) = cos (θ). Looking at the triangle and using SOHCAHTOA, we see that ![](https://media1.shmoop.com/images/calculus/cal_compderiv_morecompl_narr_latek_35.png)
![](https://media1.shmoop.com/images/calculus/cal_compderiv_morecompl_narr_graphik_3.png)
Now we can finish the simplification. ![](https://media1.shmoop.com/images/calculus/cal_compderiv_morecompl_narr_latek_36.png)
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