find the derivative of arcsin^3(x+3)

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find the derivative of arcsin^3(x+3)

Mathematics
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Okay, here's the deal, we need to look at a general right triangle with hypot. 1 and vertical leg x
This means that the horizontal leg has length \(\sqrt{1-x^2}\).
If we let \(y\) be the angle between the horizontal leg and hypot., then \[\sin y=opp/hyp=x\]

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In other words, \(\arcsin x=y\)
But then \(\cos(\arcsin x)=\cos y=adj/hyp=\sqrt{1-x^{2}}\).
Now, we are going to use the formula for derivative of inverse functions: If \(f\) and \(g\) are inverses \[g^{\prime}=1/f^{\prime}(g)\]
Let \(f(x)=\sin x\) and \(g(x)=\arcsin x\). Plugging these in gives \[(\arcsin x)^{\prime}=\frac{1}{\cos(\arcsin x)}=\frac{1}{\sqrt{1-x^{2}}}\]
using the chain rule and the definition of the derivative of arcsin I got 3[arcsinx]^2[1/(square root(1-x^2)]*[1/(square root(1-(x+3)^2)]
no clue if this is correct
Not quite, first thing is that arcsin x should be arcsin(x+3)
Okay, your main problem is that when you take the chain rule, the first multiplicand (number in the product) is \(g^{\prime}(f(x))\), that means that you need to put the whole \(f(x)\) in there, not just \(x\). So in the first chain rule, \(f(x)=arcsin(x+3)\) meaning the first part of the product is \(3*(\arcsin(x+3))^{2}\).
ok got that thanks; what about the rest of it?
In the second part, you need to replace x with x+3 and the last part is just (x+3)'=1

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