How do you implicitly differentiate #y= xy^2 + e^(x y) #?

Question

How do you implicitly differentiate #y= xy^2 +  e^(x y)   #?

Isaiah Allen · Accepted Answer

#(dy)/(dx)=(y(y+e^(xy)))/(1-2xy-xe^(xy))# In a simple function like #y=f(x)#, #y# is expressed 'explicitly' in terms of #x#. However, in many cases the variables involved in a function are not linked to each other in an explicit way and they are rather linked through an implicit formula like the one given here. In such cases, we differentiate both sides of the equality, say w.e.t. #x#, using normal formulas of differentiation such as product, quotient or chain formula, but whenever say #y# is involved, its derivative is put as #(dy)/(dx)#. The given implicit function is #y=xy^2+e^(xy)# differentiating it with respect to #x#, we get #(dy)/(dx)=y^2xx1+x xx2y(dy)/(dx)+e^(xy)xx(y+x(dy)/(dx))# or #(dy)/(dx)=y^2+2xy(dy)/(dx)+ye^(xy)+xe^(xy)(dy)/(dx)# or #(dy)/(dx)(1-2xy-xe^(xy))=y^2+ye^(xy)# or #(dy)/(dx)=(y(y+e^(xy)))/(1-2xy-xe^(xy))#

Isaiah Allen · Answer

undefined To implicitly differentiate $ y = xy^2 + e^{xy} $, we use the product rule and the chain rule.

Differentiating the term $ xy^2 $ with respect to $ x $, we get $ y^2 + 2xy\frac{dy}{dx} $.

For the term $ e^{xy} $, we use the chain rule. The derivative of $ e^{xy} $ with respect to $ x $ is $ ye^{xy} $ times $ y' $.

Combining these derivatives, we get:

$$
\frac{dy}{dx} = y^2 + 2xy\frac{dy}{dx} + ye^{xy}
$$

Rearranging terms to isolate $ \frac{dy}{dx} $, we get:

$$
\frac{dy}{dx} - 2xy\frac{dy}{dx} = y^2 + ye^{xy}
$$

$$
\frac{dy}{dx}(1 - 2xy) = y^2 + ye^{xy}
$$

$$
\frac{dy}{dx} = \frac{y^2 + ye^{xy}}{1 - 2xy}
$$