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Step-by-step Solution

Find the derivative using the quotient rule ((d/dx)((sin((x+1)^0.5*x)/(x+1)))/(((1-sin(^2)^*(x-1)*x)/(x+1))^0.5)

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Answer

$\frac{\left(\frac{1}{2}\left(x+1\right)^{-\frac{1}{2}}x+\sqrt{x+1}\right)\left(x+1\right)\cos\left(\sqrt{x+1}x\right)-\sin\left(\sqrt{x+1}x\right)}{\left(x+1\right)^2\sqrt{\frac{1+\sin\left(^2\right)^{}\left(-1\right)x^2+\sin\left(^2\right)^{}x}{x+1}}}$

Step-by-step explanation

Problem to solve:

$\frac{\frac{d}{dx}\left(\frac{sen\left(\left(\sqrt{x+1}\right)x\right)}{x+1}\right)}{\sqrt{\frac{\left(1-sen^{^2}\left(x-1\right)x\right)}{x+1}}}$
1

Multiplying polynomials $x$ and $\sin\left(^2\right)^{}\left(-1\right)x+\sin\left(^2\right)^{}$

$\frac{\frac{d}{dx}\left(\frac{\sin\left(\sqrt{x+1}x\right)}{x+1}\right)}{\sqrt{\frac{1+\sin\left(^2\right)^{}\left(-1\right)x^2+\sin\left(^2\right)^{}x}{x+1}}}$
2

Applying the quotient rule which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

$\frac{\frac{\left(x+1\right)\frac{d}{dx}\left(\sin\left(\sqrt{x+1}x\right)\right)-\sin\left(\sqrt{x+1}x\right)\frac{d}{dx}\left(x+1\right)}{\left(x+1\right)^2}}{\sqrt{\frac{1+\sin\left(^2\right)^{}\left(-1\right)x^2+\sin\left(^2\right)^{}x}{x+1}}}$

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Answer

$\frac{\left(\frac{1}{2}\left(x+1\right)^{-\frac{1}{2}}x+\sqrt{x+1}\right)\left(x+1\right)\cos\left(\sqrt{x+1}x\right)-\sin\left(\sqrt{x+1}x\right)}{\left(x+1\right)^2\sqrt{\frac{1+\sin\left(^2\right)^{}\left(-1\right)x^2+\sin\left(^2\right)^{}x}{x+1}}}$
$\frac{\frac{d}{dx}\left(\frac{sen\left(\left(\sqrt{x+1}\right)x\right)}{x+1}\right)}{\sqrt{\frac{\left(1-sen^{^2}\left(x-1\right)x\right)}{x+1}}}$

Main topic:

Differential calculus

Used formulas:

5. See formulas

Time to solve it:

~ 0.67 seconds