What is the derivative of #x = tan (x+y)#?

Question

Isaiah Allen · Accepted Answer

# dy/dx = -sin^2(x+y) #

When we differentiate #y# wrt #x# we get #dy/dx#. However, we only differentiate explicit functions of #y# wrt #x#. But if we apply the chain rule we can differentiate an implicit function of #y# wrt #y# but we must also multiply the result by #dy/dx#. Example: #d/dx(y^2) = d/dy(y^2)dy/dx = 2ydy/dx # When this is done in situ it is known as implicit differentiation. Now, we have: # x = tan(x+y) # Implicitly differentiating wrt #x# (applying product rule): # 1 = sec^2(x+y)d/dx(x+y) # # :. 1 = sec^2(x+y)(1+dy/dx ) # # :. 1+dy/dx = 1/sec^2(x+y) # # :. dy/dx = cos^2(x+y) -1 # # :. dy/dx = -sin^2(x+y) # Advanced Calculus There is another (often faster) approach using partial derivatives. Suppose we cannot find #y# explicitly as a function of #x#, only implicitly through the equation #F(x, y) = 0# which defines #y# as a function of #x, y = y(x)#. Therefore we can write #F(x, y) = 0# as #F(x, y(x)) = 0#. Differentiating both sides of this, using the partial chain rule gives us # (partial F)/(partial x) (1) + (partial F)/(partial y) dy/dx = 0 => dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y)) # So Let # F(x,y) = tan(x+y)-x #; Then; #(partial F)/(partial x) = sec^2(x+y)-1 # #(partial F)/(partial y) = sec^2(x+y) # And so: # dy/dx = -(sec^2(x+y)-1)/(sec^2(x+y)) # # " " = -(1-1/(sec^2(x+y))) # # " " = -(1-cos^2(x-y)) # # " " = -sin^2(x+y) #, as before

Christopher Agresta · Answer

undefined To find the derivative of $x = \tan(x + y)$ with respect to $x$, where $y$ is treated as a constant, use implicit differentiation. Differentiate both sides of the equation with respect to $x$:

The derivative of the left side, $x$, with respect to $x$ is $1$.

The derivative of the right side, $\tan(x + y)$, using the chain rule, is $\sec^2(x + y) \cdot \frac{d}{dx}(x + y) = \sec^2(x + y) \cdot (1 + \frac{dy}{dx})$.

Setting both derivatives equal to each other, we have:

$$1 = \sec^2(x + y) \cdot (1 + \frac{dy}{dx})$$

This equation expresses the derivative relationship, but note that it involves $\frac{dy}{dx}$, which indicates the equation is implicitly differentiated and may involve solving for $\frac{dy}{dx}$ if the context requires.