# Step-by-step Solution

## Find the derivative using the quotient rule $\frac{\frac{d}{dx}\left(\frac{\sin\left(\sqrt{x+1}x\right)}{x+1}\right)}{\sqrt{\frac{1-1\sin\left(^2\right)^{}x\left(x-1\right)}{x+1}}}$

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### Videos

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

## 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 $\sin\left(^2\right)^{}-1x+\sin\left(^2\right)^{}$

$\frac{\frac{d}{dx}\left(\frac{\sin\left(\sqrt{x+1}x\right)}{x+1}\right)}{\sqrt{\frac{1+\sin\left(^2\right)^{}-1x^2+\sin\left(^2\right)^{}x}{x+1}}}$
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 ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

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

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

Quotient rule

~ 0.88 seconds

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