Find the derivatives of the following functions: (x + y) = y2 x and f(x) = ((2x)/(1 − x2))

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Here are the derivatives for the remaining functions:

6. $\sin(x + y) = y^2 \cos x$

Differentiate both sides with respect to $x$ .
Remember to use the chain rule for terms involving $y$ and the product rule for $y^2 \cos x$ .
$\frac{d}{dx}(\sin(x + y)) = \frac{d}{dx}(y^2 \cos x)$ $\cos(x + y) \cdot \frac{d}{dx}(x + y) = \frac{d}{dx}(y^2) \cos x + y^2 \frac{d}{dx}(\cos x)$ $\cos(x + y) \left(1 + \frac{dy}{dx}\right) = 2y \frac{dy}{dx} \cos x + y^2 (-\sin x)$ $\cos(x + y) + \cos(x + y) \frac{dy}{dx} = 2y \cos x \frac{dy}{dx} - y^2 \sin x$
Rearrange the equation to group terms with $\frac{dy}{dx}$ on one side.
$\cos(x + y) \frac{dy}{dx} - 2y \cos x \frac{dy}{dx} = -y^2 \sin x - \cos(x + y)$
Factor out $\frac{dy}{dx}$ .
$\frac{dy}{dx}(\cos(x + y) - 2y \cos x) = -y^2 \sin x - \cos(x + y)$
Solve for $\frac{dy}{dx}$ .
$\frac{dy}{dx} = \frac{-y^2 \sin x - \cos(x + y)}{\cos(x + y) - 2y \cos x}$ $\frac{dy}{dx} = \boxed{\frac{y^2 \sin x + \cos(x + y)}{2y \cos x - \cos(x + y)}}$

7. $f(x) = \arctan\left(\frac{2x}{1 - x^2}\right)$
Recall the derivative of $\arctan(u)$ is $\frac{1}{1+u^2} \frac{du}{dx}$ .
Let $u = \frac{2x}{1 - x^2}$ .
Find $\frac{du}{dx}$ using the quotient rule.
Let $p = 2x$ and $q = 1 - x^2$ . Then $p' = 2$ and $q' = -2x$ .
$\frac{du}{dx} = \frac{p'q - pq'}{q^2} = \frac{(2)(1 - x^2) - (2x)(-2x)}{(1 - x^2)^2}$ $\frac{du}{dx} = \frac{2 - 2x^2 + 4x^2}{(1 - x^2)^2} = \frac{2 + 2x^2}{(1 - x^2)^2} = \frac{2(1 + x^2)}{(1 - x^2)^2}$
Substitute $u$ and $\frac{du}{dx}$ into the derivative formula for $\arctan(u)$ .
$f'(x) = \frac{1}{1 + \left(\frac{2x}{1 - x^2}\right)^2} \cdot \frac{2(1 + x^2)}{(1 - x^2)^2}$ $f'(x) = \frac{1}{1 + \frac{4x^2}{(1 - x^2)^2}} \cdot \frac{2(1 + x^2)}{(1 - x^2)^2}$ $f'(x) = \frac{1}{\frac{(1 - x^2)^2 + 4x^2}{(1 - x^2)^2}} \cdot \frac{2(1 + x^2)}{(1 - x^2)^2}$ $f'(x) = \frac{(1 - x^2)^2}{(1 - x^2)^2 + 4x^2} \cdot \frac{2(1 + x^2)}{(1 - x^2)^2}$ $f'(x) = \frac{2(1 + x^2)}{(1 - x^2)^2 + 4x^2}$
Expand the denominator.
$(1 - x^2)^2 + 4x^2 = 1 - 2x^2 + x^4 + 4x^2 = 1 + 2x^2 + x^4 = (1 + x^2)^2$
Substitute the simplified denominator back into $f'(x)$ .
$f'(x) = \frac{2(1 + x^2)}{(1 + x^2)^2}$ $f'(x) = \boxed{\frac{2}{1 + x^2}}$
Note: This result is consistent with the identity $2\arctan(x) = \arctan\left(\frac{2x}{1-x^2}\right)$ for $|x|<1$ , as $\frac{d}{dx}(2\arctan(x)) = \frac{2}{1+x^2}$ .

8. $y = \ln(\sinh x + \cosh x)$
Use the identity $\sinh x + \cosh x = e^x$ .
$y = \ln(e^x)$
Simplify the function using logarithm properties.
$y = x$
Differentiate $y$ with respect to $x$ .
$\frac{dy}{dx} = \frac{d}{dx}(x)$ $\frac{dy}{dx} = \boxed{1}$

9. $y = e^{e^x}$
Apply the chain rule.
The derivative of $e^u$ is $e^u \frac{du}{dx}$ .
Let $u = e^x$ .
$\frac{dy}{dx} = e^u \cdot \frac{du}{dx}$
Find $\frac{du}{dx}$ .
$\frac{du}{dx} = \frac{d}{dx}(e^x) = e^x$
Substitute $u$ and $\frac{du}{dx}$ back into the expression for $\frac{dy}{dx}$ .
$\frac{dy}{dx} = e^{e^x} \cdot e^x$ $\frac{dy}{dx} = \boxed{e^x e^{e^x}}$

10. $y = \frac{x}{x^2 + 1}$ (Find $\frac{d^2y}{dx^2}$ )
Find the first derivative $\frac{dy}{dx}$ using the quotient rule.
Let $u = x$ and $v = x^2 + 1$ . Then $u' = 1$ and $v' = 2x$ .
$\frac{dy}{dx} = \frac{u'v - uv'}{v^2} = \frac{(1)(x^2 + 1) - (x)(2x)}{(x^2 + 1)^2}$ $\frac{dy}{dx} = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2}$ $\frac{dy}{dx} = \frac{1 - x^2}{(x^2 + 1)^2}$
Find the second derivative $\frac{d^2y}{dx^2}$ by differentiating $\frac{dy}{dx}$ using the quotient rule again.
Let $p = 1 - x^2$ and $q = (x^2 + 1)^2$ .
Then $p' = -2x$ .
For $q'$ , use the chain rule: $q' = 2(x^2 + 1) \cdot (2x) = 4x(x^2 + 1)$ .
$\frac{d^2y}{dx^2} = \frac{p'q - pq'}{q^2}$ $\frac{d^2y}{dx^2} = \frac{(-2x)(x^2 + 1)^2 - (1 - x^2)(4x(x^2 + 1))}{((x^2 + 1)^2)^2}$ $\frac{d^2y}{dx^2} = \frac{(-2x)(x^2 + 1)^2 - 4x(1 - x^2)(x^2 + 1)}{(x^2 + 1)^4}$
Factor out common terms from the numerator, specifically $-2x(x^2 + 1)$ .
$\frac{d^2y}{dx^2} = \frac{-2x(x^2 + 1)[(x^2 + 1) + 2(1 - x^2)]}{(x^2 + 1)^4}$ $\frac{d^2y}{dx^2} = \frac{-2x[(x^2 + 1) + 2 - 2x^2]}{(x^2 + 1)^3}$ $\frac{d^2y}{dx^2} = \frac{-2x[3 - x^2]}{(x^2 + 1)^3}$ $\frac{d^2y}{dx^2} = \boxed{\frac{2x(x^2 - 3)}{(x^2 + 1)^3}}$

Find the derivatives of the following functions: (x + y) = y2 x and f(x) = ((2x)/(1 - x2))

Step-by-step solution

Differentiate both sides with respect to $x$ .

Rearrange the equation to group terms with $\frac{dy}{dx}$ on one side.

Factor out $\frac{dy}{dx}$ .

Solve for $\frac{dy}{dx}$ .

Recall the derivative of $\arctan(u)$ is $\frac{1}{1+u^2} \frac{du}{dx}$ .

Find $\frac{du}{dx}$ using the quotient rule.

Substitute $u$ and $\frac{du}{dx}$ into the derivative formula for $\arctan(u)$ .

Expand the denominator.

Substitute the simplified denominator back into $f'(x)$ .

Use the identity $\sinh x + \cosh x = e^x$ .

Simplify the function using logarithm properties.

Differentiate $y$ with respect to $x$ .

Apply the chain rule.

Find $\frac{du}{dx}$ .

Substitute $u$ and $\frac{du}{dx}$ back into the expression for $\frac{dy}{dx}$ .

Find the first derivative $\frac{dy}{dx}$ using the quotient rule.

Find the second derivative $\frac{d^2y}{dx^2}$ by differentiating $\frac{dy}{dx}$ using the quotient rule again.

Factor out common terms from the numerator, specifically $-2x(x^2 + 1)$ .

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More Mathematics Questions

Find the derivatives of the following functions: (x + y) = y2 x and f(x) = ((2x)/(1 - x2))

Step-by-step solution

Differentiate both sides with respect to xxx.

Rearrange the equation to group terms with dydx\frac{dy}{dx}dxdy​ on one side.

Factor out dydx\frac{dy}{dx}dxdy​.

Solve for dydx\frac{dy}{dx}dxdy​.

Recall the derivative of arctan⁡(u)\arctan(u)arctan(u) is 11+u2dudx\frac{1}{1+u^2} \frac{du}{dx}1+u21​dxdu​.

Find dudx\frac{du}{dx}dxdu​ using the quotient rule.

Substitute uuu and dudx\frac{du}{dx}dxdu​ into the derivative formula for arctan⁡(u)\arctan(u)arctan(u).

Expand the denominator.

Substitute the simplified denominator back into f′(x)f'(x)f′(x).

Use the identity sinh⁡x+cosh⁡x=ex\sinh x + \cosh x = e^xsinhx+coshx=ex.

Simplify the function using logarithm properties.

Differentiate yyy with respect to xxx.

Apply the chain rule.

Find dudx\frac{du}{dx}dxdu​.

Substitute uuu and dudx\frac{du}{dx}dxdu​ back into the expression for dydx\frac{dy}{dx}dxdy​.

Find the first derivative dydx\frac{dy}{dx}dxdy​ using the quotient rule.

Find the second derivative d2ydx2\frac{d^2y}{dx^2}dx2d2y​ by differentiating dydx\frac{dy}{dx}dxdy​ using the quotient rule again.

Factor out common terms from the numerator, specifically −2x(x2+1)-2x(x^2 + 1)−2x(x2+1).

Still stuck on this one?

More Mathematics Questions

Differentiate both sides with respect to $x$ .

Rearrange the equation to group terms with $\frac{dy}{dx}$ on one side.

Factor out $\frac{dy}{dx}$ .

Solve for $\frac{dy}{dx}$ .

Recall the derivative of $\arctan(u)$ is $\frac{1}{1+u^2} \frac{du}{dx}$ .

Find $\frac{du}{dx}$ using the quotient rule.

Substitute $u$ and $\frac{du}{dx}$ into the derivative formula for $\arctan(u)$ .

Substitute the simplified denominator back into $f'(x)$ .

Use the identity $\sinh x + \cosh x = e^x$ .

Differentiate $y$ with respect to $x$ .

Find $\frac{du}{dx}$ .

Substitute $u$ and $\frac{du}{dx}$ back into the expression for $\frac{dy}{dx}$ .

Find the first derivative $\frac{dy}{dx}$ using the quotient rule.

Find the second derivative $\frac{d^2y}{dx^2}$ by differentiating $\frac{dy}{dx}$ using the quotient rule again.

Factor out common terms from the numerator, specifically $-2x(x^2 + 1)$ .