Question

***URGENT***

Find the derivative of y with respect to x if 2^(xy) = 1. Show all your work.

Answer 1

Answer:

$\sf \dfrac{d(y)}{dx} = -\dfrac{y}{x}$

Step-by-step explanation:

Let's find the derivative of $\sf y$ with respect to $\sf x$ in the equation $\sf 2^{xy} = 1$. We'll use logarithmic differentiation to solve this problem.

Starting with the given equation:

$\sf 2^{xy} = 1$

Take the natural logarithm (ln) of both sides:

$\sf \ln(2^{xy}) = \ln(1)$

Apply the power rule and the logarithm law:$\sf ln(x^a) = a \ln(x)$ and $\sf \ln(x) = \dfrac{1}{x}$

$\sf xy \ln(2) = 0$

Now, differentiate both sides with respect to $\sf x$, applying the product rule:

$\sf \dfrac{d}{dx}(xy \ln(2)) = \dfrac{d}{dx}(0)$

Use the product rule $\sf \dfrac{d}{dx}(uv) = u'\cdot v + u \cdot v'$:

$\sf y \ln(2) + x \cdot \dfrac{d(y)}{dx} \cdot \ln(2) = 0$

Now, isolate $\sf \dfrac{dy}{dx}$:

$\sf x \cdot \dfrac{d(y)}{dx} \cdot \ln(2) = -y \ln(2)$

Divide both sides by $\sf x \ln(2)$:

$\sf \dfrac{d(y)}{dx} = -\dfrac{y}{x}$

So, the derivative of $\sf y$ with respect to $\sf x$ is $\sf -\dfrac{y}{x}$.

Answer 2

Answer:

$\dfrac{\text{d}y}{\text{d}x}=-\dfrac{y}{x}$

Step-by-step explanation:

To find the derivative of y with respect to x given the equation $2^{xy} = 1$, we can use logarithmic differentiation.

First, take the natural logarithm (ln) of both sides of the equation:

$\ln(2^{xy}) = \ln(1)$

Now, apply the power rule to simplify the left side:

$xy \ln(2) = \ln(1)$

Since ln(1) = 0, we get:

$xy \ln(2) = 0$

Now, differentiate both sides with respect to x:

$\dfrac{\text{d}}{\text{d}x}(xy \ln(2)) = \dfrac{\text{d}}{\text{d}x}(0)$

Apply the product rule of differentiation on the left side.

$\boxed{\begin{array}{c}\underline{\sf Product\;Rule\;for\;Differentiation}\\\\\textsf{If}\;y=uv\;\textsf{then:}\\\\\dfrac{\text{d}y}{\text{d}x}=u\dfrac{\text{d}v}{\text{d}x}+v\dfrac{\text{d}u}{\text{d}x}\end{array}}$

$\textsf{Let}\;\;u=x \implies \dfrac{\text{d}u}{\text{d}x}=1$

$\textsf{Let}\;\;v=y\ln(2) \implies \dfrac{\text{d}v}{\text{d}x}=\ln(2)\dfrac{\text{d}y}{\text{d}x}$

Therefore:

$\begin{aligned}\dfrac{\text{d}}{\text{d}x}(xy \ln(2)) &= x\cdot \ln(2)\dfrac{\text{d}y}{\text{d}x}+y\ln(2) \cdot 1\\\\&=x\ln(2)\dfrac{\text{d}y}{\text{d}x}+y\ln(2)\end{aligned}$

So the differentiated equation is:

$x\ln(2)\dfrac{\text{d}y}{\text{d}x}+y\ln(2)=0$

Now, solve for dy/dx:

$x\ln(2)\dfrac{\text{d}y}{\text{d}x}=-y\ln(2)$

$\dfrac{\text{d}y}{\text{d}x}=\dfrac{-y\ln(2)}{x\ln(2)}$

$\dfrac{\text{d}y}{\text{d}x}=-\dfrac{y}{x}$

So, the derivative of y with respect to x if $2^{xy} = 1$ is:

$\Large\boxed{\boxed{\dfrac{\text{d}y}{\text{d}x}=-\dfrac{y}{x}}}$