How do you find #dy/dx# by implicit differentiation of #x^2-y^2=16#?

Question

Evelyn Agard · Accepted Answer

# dy/dx = x/y # If an equation does not express #y# explicitly in terms of #x# as in #y=f(x)#, and instead we have #g(y)=f(x)#, then when we differentiate we apply the chain rule so that we differentiate #g(y)# wrt #y# rather than #x# as in: # g(y) = f(x) # # :. d/dx g(y) = d/dx f(x) # # :. dy/dx d/dy g(y) = f'(x) # # :. g'(y) dy/dx= f'(x) # This will typically result in #dy/dx# being a function of #x# and #y#, rather than #dy/dx# being a function of #x# alone as we usually get So for # x^2 - y^2 = 16 # we have: # d/dx(x^2) - d/dx(y^2) = d/dx(16) # # :. 2x - dy/dxd/dy(y^2) = 0 # # :. 2x - 2ydy/dx = 0 # # :. 2ydy/dx = 2x # # :. dy/dx = x/y #

William Abdalla · Answer

undefined To find $ \frac{dy}{dx} $ by implicit differentiation of $ x^2 - y^2 = 16 $, differentiate both sides of the equation with respect to $ x $, treating $ y $ as a function of $ x $ using the chain rule. Then solve for $ \frac{dy}{dx} $.

Differentiating both sides with respect to $ x $, we get:

$$ \frac{d}{dx}(x^2) - \frac{d}{dx}(y^2) = \frac{d}{dx}(16) $$

$$ 2x - \frac{d}{dx}(y^2) = 0 $$

$$ 2x - 2y\frac{dy}{dx} = 0 $$

Now, solve for $ \frac{dy}{dx} $:

$$ 2y\frac{dy}{dx} = 2x $$

$$ \frac{dy}{dx} = \frac{2x}{2y} $$

$$ \frac{dy}{dx} = \frac{x}{y} $$

Thus, $ \frac{dy}{dx} = \frac{x}{y} $.