Trigonometric Implicit Relations

Why Trigonometric Relations Need Careful Differentiation

A relation may involve \(x\) and \(y\) without giving \(y\) explicitly as a function of \(x\). If it contains trigonometric expressions as well, differentiation requires two ideas at once: the usual derivatives of sine and cosine, and the Chain Rule for their inputs.

For example, in \(\sin(y)+\cos(xy)=x\), the quantity \(y\) can vary with \(x\), and the product \(xy\) can vary with \(x\) as well. The derivative of \(\sin(y)\) therefore contains \(\frac{dy}{dx}\), while the derivative of \(\cos(xy)\) involves the derivative of the entire product \(xy\).

Throughout, angles are measured in radians, as required for the standard trigonometric derivative formulas. We seek the slope of a differentiable local branch \(y=y(x)\) passing through a point on the relation.

The Differentiation Rule

Each trigonometric derivative has two parts: the derivative of the outer trigonometric function and the derivative of its input. In implicit differentiation, that inner derivative may itself contain \(y'\). The main algebraic task is then to collect all terms involving \(y'\) on one side.

Deriving the Slope Formula

Consider the relation

Suppose a differentiable branch \(y=y(x)\) satisfies this relation near the point of interest. Differentiate both sides with respect to \(x\). For the first term, the Chain Rule gives \(\cos(y)y'\). For the second, first differentiate the product \(xy\), obtaining \(y+xy'\), and then apply the cosine derivative.

Expand the product and gather the terms containing \(y'\):

If \(\cos(y)-x\sin(xy)\ne 0\) at the point, division gives the slope there:

Why the Formula Works

The derivation is an instance of the general rule for differentiating a relation. If \(F(x,y)=0\) and \(y\) is a differentiable function of \(x\), then differentiating the composite expression \(F(x,y(x))\) gives

where \(F_x\) and \(F_y\) denote the partial derivatives with respect to the indicated variables. This equation follows from the multivariable Chain Rule: as \(x\) changes, \(F\) changes both because its first input changes and because its second input \(y(x)\) changes. When \(F_y\ne 0\), solving the equation yields \(y'=-F_x/F_y\).

For the relation above, take \(F(x,y)=\sin(y)+\cos(xy)-x\). Its partial derivatives are \(F_x=-y\sin(xy)-1\) and \(F_y=\cos(y)-x\sin(xy)\). Thus \(-F_x/F_y\) is exactly the formula derived by differentiating term by term. The nonzero-denominator condition is the requirement \(F_y\ne 0\) for this slope calculation.

Worked Examples

A Concrete Chain-Rule Pitfall

A common mistake is to differentiate \(\cos(xy)\) as though its input were just \(x\), writing \(-\sin(xy)\). That omits the derivative of \(xy\). The correct derivative along a branch is

The omitted factor matters even when the product input is complicated. For example, replacing the correct term by \(-\sin(xy)\) in the first worked relation would give \(\cos(y)y'-\sin(xy)=1\). At \((1,0)\), that incorrect equation becomes \(y'=1\), which happens to give the same numerical slope because \(\sin(0)=0\). This accidental agreement does not validate the calculation: away from that point, the missing \(y+xy'\) factor changes the equation and generally changes the result.

A Reliable Working Procedure

For a trigonometric relation, keep the differentiation and point evaluation in a clear order:

Why This Skill Matters

Implicit relations can describe curves whose equations are difficult or impossible to rearrange conveniently as \(y=f(x)\). Trigonometric terms add nested inputs, but the underlying method remains systematic: differentiate the full expression, account for every changing input, and solve the resulting equation for the slope when its coefficient permits.

The most important check is not merely whether the final fraction looks plausible. Confirm that the point satisfies the original relation, that every composite input received its own derivative, and that the proposed slope satisfies the differentiated equation at the point.