Question

y=ln(arcsinx)(x2-5x)1/2

Answer 1

What about it?

Answer 2

am asked to derive

Answer 3

Okay. So is it:
\[y = \ln(\sin^{-1}(x))\sqrt{x^2+5x}\]?

Answer 4

ln(arcsinx)

Answer 5

yeah that's the same as what I have. So we start by noting the chain rule and product rule will both rear their heads.

Answer 6

ok

Answer 7

Therefore we need:
\[\frac{d}{dx}(fg) = f'g + fg'; \frac{d}{dx}f(g(x)) = f'(g(x)) g'(x)\]
So we have:
\[f(x) = \ln(\arcsin(x)); g(x) = \sqrt{x^2-5x}\]
Applying the product rule:
\[\frac{d}{dx}f(x)g(x) = (\ln(\arcsin(x))'*\sqrt{x^2-5x} + \ln(\arcsin(x)) *(\sqrt{x^2-5x})'\]
So what is:
\[\frac{d}{dx}\ln(\arcsin(x)); \text{and}; \frac{d}{dx}\sqrt{x^2-5x} \text{ ?}\]

Answer 8

(1/arcsinx)(1/(1-x2)......1

(2x-5).1/2(x2-5x)-1/2........2

Answer 9

thank you very much

Answer 10

\[ \frac{ 1 }{ arcosin(x) }\frac{ 1 }{ \sqrt{1-x^2} }...1\]

\[\frac{ 2x-5 }{ 2\sqrt{x^2-5x} }...2\]

Answer 11

Then do the substitutions as he said and that's it

Answer 12

thank you sir