What is the implicit derivative of #25=cosy/x-3xy#?

Question

Scarlett Allison · Accepted Answer

#dy/dx = -(25 +6xy)/(siny+3x^2)#

Differentiate both sides of the equation with respect to #x#:

#d/dx(cosy/x-3xy) = 0#

# (-xsinydy/dx-cosy)/x^2 -3(y+xdy/dx) = 0#

Since #x != 0# we can multiply both sides by #x#:

# -sinydy/dx-cosy/x -3xy-3x^2dy/dx = 0#

#dy/dx(siny+3x^2) = -cosy/x -3xy#

From the original expression we can substitute:

#cosy/x = 25+3xy#

so:

#dy/dx(siny+3x^2) = -25 -6xy#

and finally:

#dy/dx = -(25 +6xy)/(siny+3x^2)#

Emma Aldridge · Answer

undefined To find the implicit derivative of $25 = \frac{\cos y}{x} - 3xy$, we will differentiate both sides of the equation with respect to $x$.

Given equation: $25 = \frac{\cos y}{x} - 3xy$

Taking the derivative of both sides with respect to $x$:

$\frac{d}{dx}(25) = \frac{d}{dx}\left(\frac{\cos y}{x} - 3xy\right)$

Since $25$ is a constant, its derivative is $0$.

Now, using the quotient rule for differentiation on $\frac{\cos y}{x}$:

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

Applying the chain rule to $\frac{d}{dx}(\cos y)$:

$\frac{d}{dx}(\cos y) = -\sin y \cdot \frac{dy}{dx}$

Substituting this back into the quotient rule:

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

Now, differentiating $3xy$ with respect to $x$ using the product rule:

$\frac{d}{dx}(3xy) = 3y + 3x \frac{dy}{dx}$

Now, putting it all together:

$$0 = -\frac{\cos y}{x^2} - \frac{x \cdot (-\sin y \cdot \frac{dy}{dx})}{x^2} - 3y - 3x \frac{dy}{dx}$$

$$0 = -\frac{\cos y}{x^2} + \frac{\sin y}{x} \frac{dy}{dx} - 3y - 3x \frac{dy}{dx}$$

Rearranging terms:

$$\frac{\sin y}{x} \frac{dy}{dx} - 3x \frac{dy}{dx} = 3y + \frac{\cos y}{x^2}$$

Factoring out $ \frac{dy}{dx} $:

$$\frac{dy}{dx}(\frac{\sin y}{x} - 3x) = 3y + \frac{\cos y}{x^2}$$

Finally, solving for $ \frac{dy}{dx} $:

$$\frac{dy}{dx} = \frac{3y + \frac{\cos y}{x^2}}{\frac{\sin y}{x} - 3x}$$

Hence, the implicit derivative of $25 = \frac{\cos y}{x} - 3xy$ with respect to $x$ is:

$$\frac{dy}{dx} = \frac{3y + \frac{\cos y}{x^2}}{\frac{\sin y}{x} - 3x}$$