Question

Differentiate implicitly to find the first partial derivatives of \(z\) \(\tan (x+y)+\tan (y+z)=1\)

Short answer

The first partial derivative of z with respect to x is \(y' = -\frac{1}{1+\tan^2(x+y)}\), the first partial derivative of z with respect to y is \(z' = -\frac{1+\tan^2(x+y)}{1+ \tan^2(y+z)}\), and the equation \(\tan^2(y+z) = -1\) must be satisfied.

Step-by-step solution

Differentiating with respect to x

Differentiating the entire equation with respect to x, treating y and z as functions of x. Application of the chain rule gives: \((1+\tan^2(x+y))y'+(1+\tan^2(y+z))0 = 0\), which simplifies to \(y' = -\frac{1}{1+\tan^2(x+y)}\).

Differentiating with respect to y

Differentiating the entire equation with respect to y, considering z as a function of y. Application of the chain rule gives: \((1+\tan^2(x+y))+(1+\tan^2(y+z))z'+(1+\tan^2(y+z)) = 0\), which simplifies to \(z' = -\frac{1+\tan^2(x+y)}{1+ \tan^2(y+z)}\).

Differentiating with respect to z

Differentiating the entire equation with respect to z, treating x and y as constants: \((1+\tan^2(x+y))0+(1+\tan^2(y+z)) = 0\), which simplifies to \(0 = -1-\tan^2(y+z)\) and therefore \(\tan^2(y+z) = -1\).

Key concepts

Partial Derivatives

When dealing with functions of multiple variables, partial derivatives are the tool we use to explore how a function changes in response to variations in just one of its inputs, while keeping others fixed. Think of it like focusing your attention on one variable at a time.

In the textbook exercise, we have a function involving both x and y. When finding the partial derivative with respect to x, denoted as \(y'\) or \(\frac{\partial z}{\partial x}\), y is treated like it might depend on x, which introduces the need for implicit differentiation. This means we don't solve for z explicitly but differentiate as if z is some function of both x and y.

The secret sauce here is the product rule from single-variable calculus: when a term involves a product of functions (like x and y), each part gets to have its turn being differentiated. However, because z is 'partially' hitched to both x and y, we'll use the chain rule to account for the indirect effect via y—which leads nicely into our next concept.

Chain Rule

The chain rule is a real superstar in calculus. Simply put, it deals with how to differentiate a function of a function. When moving down the differentiation road and you hit a composite function, you don't stop; you apply the chain rule! For instance, if we're looking at \(\tan(x+y)\), and we need to differentiate with respect to x, the chain rule directs us to first differentiate \(\tan(u)\) with respect to its inner function u (where u = x + y), and then multiply by the derivative of u with respect to x. In our exercise, this is why we see terms like \(1+\tan^2(x+y)\) appearing—they come from differentiating \(\tan(u)\) with respect to u and then multiplying by the derivative of u with respect to x, which is 1 for x and y' for y (since y is treated as a function of x).

This handshaking between layers of functions is what makes the chain rule so essential in dealing with composite functions. It's especially crucial in implicit differentiation where the relationships between variables are intertwined.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent are periodic and oscillate between specific values. In calculus, we not only study their properties but also how they change—that's where derivatives come in. In our exercise, the focus is on the tangent function, notated as \(\tan\). The derivative of \(\tan(x)\) with respect to x is \(1+\tan^2(x)\), based on the identity \(1+\tan^2(x) = \sec^2(x)\). When differentiating an equation involving \(\tan\) of some expression, like \(\tan(x+y)\), we use this derivative but remember to employ the chain rule since we're now differentiating with respect to a function, not just a single variable.

A good reminder for students: always watch out for the identities and derivative rules related to trigonometric functions—they often pop up in calculus problems and can be a sneaky detail if forgotten. And in our exercise, seeing an equation that says \(\tan^2(y+z) = -1\) should raise some red flags since the square of tangent, based on its fundamental properties, can never be negative. This leads to an inconsistency, indicative of a possible mistake or an unattainable condition within the context of the problem.