Apply the quotient rule, d/dx(u/v) = (u'v - uv')/(v²), where u = x² and v = x+1.

Question

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Here are the solutions for the differentiation problems:

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

Apply the quotient rule, $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$ , where $u = x^2$ and $v = x+1$ .

Find the derivatives of $u$ and $v$ .

$u' = \frac{d}{dx}(x^2) = 2x$  
$v' = \frac{d}{dx}(x+1) = 1$

Substitute $u, v, u', v'$ into the quotient rule formula.

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

Simplify the expression.

$$
\frac{d}{dx}\left(\frac{x^2}{x+1}\right) = \frac{2x^2 + 2x - x^2}{(x+1)^2}
$$
$$
\frac{d}{dx}\left(\frac{x^2}{x+1}\right) = \frac{x^2 + 2x}{(x+1)^2}
$$
The derivative is $\boxed{\frac{x^2 + 2x}{(x+1)^2}}$.

2. $\frac{d}{dx}\left(\frac{\sin x}{x}\right)$

Apply the quotient rule, $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$ , where $u = \sin x$ and $v = x$ .

Find the derivatives of $u$ and $v$ .

$u' = \frac{d}{dx}(\sin x) = \cos x$  
$v' = \frac{d}{dx}(x) = 1$

Substitute $u, v, u', v'$ into the quotient rule formula.

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

Simplify the expression.

$$
\frac{d}{dx}\left(\frac{\sin x}{x}\right) = \frac{x \cos x - \sin x}{x^2}
$$
The derivative is $\boxed{\frac{x \cos x - \sin x}{x^2}}$.

3. $\frac{d}{dx}\left(\frac{x^3}{x^2+1}\right)$

Apply the quotient rule, $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$ , where $u = x^3$ and $v = x^2+1$ .

Find the derivatives of $u$ and $v$ .

$u' = \frac{d}{dx}(x^3) = 3x^2$  
$v' = \frac{d}{dx}(x^2+1) = 2x$

Substitute $u, v, u', v'$ into the quotient rule formula.

$$
\frac{d}{dx}\left(\frac{x^3}{x^2+1}\right) = \frac{(3x^2)(x^2+1) - (x^3)(2x)}{(x^2+1)^2}
$$

Simplify the expression.

$$
\frac{d}{dx}\left(\frac{x^3}{x^2+1}\right) = \frac{3x^4 + 3x^2 - 2x^4}{(x^2+1)^2}
$$
$$
\frac{d}{dx}\left(\frac{x^3}{x^2+1}\right) = \frac{x^4 + 3x^2}{(x^2+1)^2}
$$
The derivative is $\boxed{\frac{x^4 + 3x^2}{(x^2+1)^2}}$.

4. $\frac{d}{dx}\left(\frac{\ln x}{x}\right)$

Apply the quotient rule, $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$ , where $u = \ln x$ and $v = x$ .

Find the derivatives of $u$ and $v$ .

$u' = \frac{d}{dx}(\ln x) = \frac{1}{x}$  
$v' = \frac{d}{dx}(x) = 1$

Substitute $u, v, u', v'$ into the quotient rule formula.

$$
\frac{d}{dx}\left(\frac{\ln x}{x}\right) = \frac{\left(\frac{1}{x}\right)(x) - (\ln x)(1)}{x^2}
$$

Simplify the expression.

$$
\frac{d}{dx}\left(\frac{\ln x}{x}\right) = \frac{1 - \ln x}{x^2}
$$
The derivative is $\boxed{\frac{1 - \ln x}{x^2}}$.

Apply the quotient rule, (d)/(dx)((u)/(v)) = (u'v - uv')/(v²), where u = x² and v = x+1.

Step-by-step solution

Apply the quotient rule, $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$ , where $u = x^2$ and $v = x+1$ .

Find the derivatives of $u$ and $v$ .

Substitute $u, v, u', v'$ into the quotient rule formula.

Simplify the expression.

Apply the quotient rule, $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$ , where $u = \sin x$ and $v = x$ .

Find the derivatives of $u$ and $v$ .

Substitute $u, v, u', v'$ into the quotient rule formula.

Simplify the expression.

Apply the quotient rule, $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$ , where $u = x^3$ and $v = x^2+1$ .

Find the derivatives of $u$ and $v$ .

Substitute $u, v, u', v'$ into the quotient rule formula.

Simplify the expression.

Apply the quotient rule, $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$ , where $u = \ln x$ and $v = x$ .

Find the derivatives of $u$ and $v$ .

Substitute $u, v, u', v'$ into the quotient rule formula.

Simplify the expression.

Still stuck on this one?

More Mathematics Questions

Apply the quotient rule, (d)/(dx)((u)/(v)) = (u'v - uv')/(v²), where u = x² and v = x+1.

Step-by-step solution

Apply the quotient rule, ddx(uv)=u′v−uv′v2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}dxd​(vu​)=v2u′v−uv′​, where u=x2u = x^2u=x2 and v=x+1v = x+1v=x+1.

Find the derivatives of uuu and vvv.

Substitute $u, v, u', v'$ into the quotient rule formula.

Simplify the expression.

Apply the quotient rule, ddx(uv)=u′v−uv′v2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}dxd​(vu​)=v2u′v−uv′​, where u=sin⁡xu = \sin xu=sinx and v=xv = xv=x.

Find the derivatives of uuu and vvv.

Substitute $u, v, u', v'$ into the quotient rule formula.

Simplify the expression.

Apply the quotient rule, ddx(uv)=u′v−uv′v2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}dxd​(vu​)=v2u′v−uv′​, where u=x3u = x^3u=x3 and v=x2+1v = x^2+1v=x2+1.

Find the derivatives of uuu and vvv.

Substitute $u, v, u', v'$ into the quotient rule formula.

Simplify the expression.

Apply the quotient rule, ddx(uv)=u′v−uv′v2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}dxd​(vu​)=v2u′v−uv′​, where u=ln⁡xu = \ln xu=lnx and v=xv = xv=x.

Find the derivatives of uuu and vvv.

Substitute $u, v, u', v'$ into the quotient rule formula.

Simplify the expression.

Still stuck on this one?

More Mathematics Questions

Apply the quotient rule, $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$ , where $u = x^2$ and $v = x+1$ .

Find the derivatives of $u$ and $v$ .

Apply the quotient rule, $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$ , where $u = \sin x$ and $v = x$ .

Find the derivatives of $u$ and $v$ .

Apply the quotient rule, $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$ , where $u = x^3$ and $v = x^2+1$ .

Find the derivatives of $u$ and $v$ .

Apply the quotient rule, $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$ , where $u = \ln x$ and $v = x$ .

Find the derivatives of $u$ and $v$ .