Why Evaluate the Slope Directly?
An equation such as \(x^2+y^2=25\) describes a curve, but it does not give one single value of \(y\) for each \(x\) across the whole curve. Solving for \(y\) would produce two expressions, \(y=\sqrt{25-x^2}\) and \(y=-\sqrt{25-x^2}\), each describing only part of it.
If the goal is to find the slope at one specified point, solving for \(y\) first is often unnecessary. Instead, differentiate the original relation, keep the derivative \(dy/dx\) in the equation, and then substitute the point’s coordinates. The result is usually a numerical equation that can be solved directly for the slope.
Differentiate First, Substitute Second
Suppose an equation relates \(x\) and \(y\), and near a point of interest it describes a differentiable branch \(y=y(x)\). In the differentiation, \(y\) depends on \(x\). Thus, for example, the chain rule gives \(\frac{d}{dx}(y^2)=2y\frac{dy}{dx}\), not just \(2y\).
After differentiating, substitute the specified values \(x=x_0\) and \(y=y_0\). The remaining unknown is \(dy/dx\) at that point. The point must satisfy the original relation, and the resulting equation must determine a unique value for the slope.
Why Substitution Works
The justification is that implicit differentiation produces an equation that holds along the differentiable branch, not just at one selected point. The chain rule accounts for how \(y\) changes with \(x\). Evaluating that equation at a point on the branch therefore gives an equation involving the derivative at that point.
In particular, suppose differentiating a relation and collecting terms gives an equation of the form \(A(x,y)\frac{dy}{dx}+B(x,y)=0\). At a point \((x_0,y_0)\) on the curve, substitution gives \(A(x_0,y_0)\frac{dy}{dx}+B(x_0,y_0)=0\). If \(A(x_0,y_0)\ne 0\), there is a unique slope:
This conclusion requires a differentiable branch through the point and a nonzero coefficient of \(dy/dx\) in the differentiated equation. If that coefficient is zero, this particular equation does not determine a unique finite slope; one should not divide by zero or claim a value for \(dy/dx\) from it.
Worked Examples
Worked Example: Slope on a Circle
Find the slope of \(x^2+y^2=25\) at \((3,4)\), without solving the original equation for \(y\).
First check that the point lies on the curve:
Differentiate both sides with respect to \(x\). Since \(y\) depends on \(x\), differentiating \(y^2\) gives \(2y\frac{dy}{dx}\):
Now substitute \(x=3\) and \(y=4\):
Subtract \(6\) from both sides, then divide by \(8\):
Check by substituting this value into the differentiated equation at the point:
The point also satisfies the original equation, as checked above. Therefore the slope at \((3,4)\) is \(-\frac{3}{4}\).
Worked Example: A Relation with an \(xy\) Term
Find the slope of \(x^2+xy+y^2=7\) at \((1,2)\).
Verify the point in the original relation:
Differentiate each term. The product rule gives \(\frac{d}{dx}(xy)=y+x\frac{dy}{dx}\), and the chain rule gives \(\frac{d}{dx}(y^2)=2y\frac{dy}{dx}\). Therefore,
Substitute \(x=1\) and \(y=2\):
Subtract \(4\), then divide by \(5\):
Check in the differentiated equation at the specified point:
The point lies on the original curve, and the derivative equation is satisfied by \(-\frac45\). Thus the slope at \((1,2)\) is \(-\frac45\).
Worked Example: Differentiating a Composite Expression
Find the slope at \((1,1)\) on the curve \((x+y)^3+xy=9\).
First check the point:
Differentiate the first term using the chain rule. The derivative of \(x+y\) is \(1+\frac{dy}{dx}\). Differentiate the product \(xy\) using the product rule:
At \((1,1)\), \(x+y=2\), so \((x+y)^2=4\). Substituting the coordinates gives:
Expand and collect the derivative terms:
Subtract \(13\) and divide by \(13\):
Check the value in the differentiated equation after substitution:
The original relation is also satisfied at \((1,1)\), as verified above. Hence the slope is \(-1\).
A Reliable Order of Operations
The order matters. Differentiate the relation before substituting the point. If you substitute too early, you may turn the equation into a numerical statement and lose the information needed to determine how \(y\) changes as \(x\) changes.
Substitute its coordinates into the original relation. The point must lie on the curve.
Treat \(y\) as a function of \(x\), applying the chain rule and product rule wherever needed.
Substitute the given \(x\)- and \(y\)-coordinates into the differentiated equation.
Find the value of \(dy/dx\), then substitute it back into the evaluated derivative equation.
Common Mistake: Treating \(y\) as a Constant
A frequent error is to differentiate a term involving \(y\) as though \(y\) were constant. For example, an incorrect differentiation of \(x^2+y^2=25\) is \(2x+2y=0\). This omits the factor \(dy/dx\) from the chain rule.
At \((3,4)\), that incorrect equation becomes:
which is not \(0\). It cannot give a valid slope. The correction is to differentiate \(y^2\) as \(2y\frac{dy}{dx}\), giving \(2x+2y\frac{dy}{dx}=0\). Substitution at \((3,4)\) then gives \(6+8\frac{dy}{dx}=0\), and therefore \(\frac{dy}{dx}=-\frac34\).
When the Method Does Not Give a Slope
The method depends on two checks: the point must be on the original curve, and the evaluated differentiated equation must determine a unique value of \(dy/dx\). In the collected form \(A(x,y)\frac{dy}{dx}+B(x,y)=0\), if \(A(x_0,y_0)\ne 0\), division gives one slope. If \(A(x_0,y_0)=0\), this equation alone cannot be divided through to find a unique finite slope.
That failure does not justify assigning a slope by guesswork. It means the stated calculation does not determine one. Questions about points where the usual slope calculation fails, and how to classify the resulting curve behavior, require separate analysis.
Why This Technique Is Useful
An implicit equation can describe a curve without providing a convenient formula for \(y\). Even if solving for \(y\) is possible, it may produce several branches and make the differentiation harder. When only the slope at one known point is needed, direct substitution into the differentiated relation avoids that detour.
The method also makes the role of the point explicit: its coordinates turn a derivative relation involving \(x\), \(y\), and \(dy/dx\) into a numerical equation. The essential habits are to keep the chain rule factors, verify the point, and check that the resulting equation really determines the requested derivative.
Check Your Understanding
For each question, differentiate before substituting the specified coordinates.
- Verify that \((0,5)\) lies on \(x^2+y^2=25\), then find the slope there.
- For \(x^2+xy+y^2=7\), find the slope at \((2,1)\).
- For \((x+y)^2=16\), find the slope at \((1,3)\).
- In the equation \(x^2+y^2=25\), what chain-rule factor must accompany the derivative of \(y^2\)?
- In the collected equation \(A(x,y)\frac{dy}{dx}+B(x,y)=0\), what condition allows you to solve uniquely for \(dy/dx\) at a specified point?
