Why Exponential and Logarithmic Relations Need Care
An equation involving \(x\) and \(y\) may define a curve without expressing \(y\) as a formula in terms of \(x\). When a curve can locally be written as \(y=y(x)\), differentiating its equation requires treating every occurrence of \(y\) as a function of \(x\).
This matters especially for exponentials and logarithms. Differentiating \(e^y\) requires a chain-rule factor of \(y'\), and differentiating \(\ln(x+y)\) requires the derivative of the entire argument \(x+y\). In a product such as \(xy\), both factors contribute.
The Differentiation Rules Behind the Method
Suppose \(y=y(x)\). By the chain rule,
The factor \(y'\) appears because the exponent \(y\) changes as \(x\) changes. Similarly, if \(x+y>0\), then
For the product \(xy\), the product rule gives
These rules are standard differentiation rules applied while remembering the dependence \(y=y(x)\). The main task is then to collect the terms containing \(y'\) and solve for the slope.
A General Derivation for Exponential and Logarithmic Relations
Consider first the relation
Assume that near the point of interest the curve defines a differentiable function \(y=y(x)\). Differentiate each term with respect to \(x\). The constant on the right has derivative zero:
The terms involving \(y'\) can be grouped:
Therefore, whenever \(e^y+x\ne 0\), the slope is
The condition \(e^y+x\ne0\) is needed to divide by the coefficient of \(y'\). If it fails, this formula does not determine the slope; that does not by itself prove that no tangent exists.
Now consider
The logarithm requires \(x+y>0\). Differentiate both sides, applying the chain rule to the logarithm and to \(y^2\):
Multiplying by \(x+y\), which is positive on the domain, gives
Move the terms involving \(y'\) to one side:
Consequently, when \(2y(x+y)-1\ne0\),
The formula requires both that the point lie in the logarithm’s domain and that its denominator be nonzero.
Worked Examples
Worked Example: Tangent to an Exponential Relation
Find the tangent line to \(e^y+xy=1\) at \((1-e,1)\).
First check that the point lies on the curve. At \(x=1-e\) and \(y=1\),
Differentiate the relation:
Collect the \(y'\) terms and solve:
At the specified point, \(e^y=e\), so the denominator is
Thus the slope is
Using the point-slope form of a line, the tangent is
Check: the line gives \(y=2-e-(1-e)=1\) at \(x=1-e\), and its slope is \(-1\), as calculated. The point also satisfies the original relation, as verified above.
Worked Example: A Logarithmic Relation at a Point with \(y=0\)
Find the tangent line to \(\ln(x+y)=y^2\) at \((1,0)\).
The point is in the logarithm’s domain because \(x+y=1>0\). It lies on the curve because
Differentiate both sides, keeping the derivative of \(x+y\) inside the logarithm:
At \((1,0)\), this becomes
The tangent line through \((1,0)\) with slope \(-1\) is
Check: at \(x=1\), this line gives \(y=0\), and its slope is \(-1\). The original relation holds at the point because \(\ln(1)=0=0^2\).
Worked Example: A Logarithmic Relation at a Point with \(y=1\)
Find the slope and tangent line to \(\ln(x+y)=y^2\) at \((e-1,1)\).
Check the point and the logarithm’s domain:
From implicit differentiation,
provided the denominator is nonzero. At the point, that denominator is
Since \(e>1\), \(2e-1>1\), so division is valid. The slope is
The tangent line is therefore
Check: substituting \(x=e-1\) gives \(y=1\), and the line’s slope is \(1/(2e-1)\). The original relation holds at the point because \(\ln(e)=1=1^2\).
A Common Mistake: Losing the Dependent-Variable Factor
A frequent error is to differentiate \(e^y\) as though \(y\) were the independent variable. That omits the chain-rule factor \(y'\). For example, the incorrect calculation
would lead to the wrong differentiated equation
The correction is to use \(y=y(x)\):
The same issue arises with logarithms. The derivative of \(\ln(x+y)\) is not simply \(1/(x+y)\), because its argument changes with \(x\). The correct expression is
A useful check is to differentiate the inside expression explicitly before applying the outer derivative rule. For \(\ln(x+y)\), the inside derivative is \(1+y'\), not just \(1\).
A Reliable Calculation Process
Substitute the point into the original relation. For a logarithm, also verify that its argument is positive.
Use the chain rule for exponentials and logarithms, and the product rule for products such as \(xy\).
Gather all terms containing \(y'\) on one side and isolate them from terms without \(y'\).
Before dividing, confirm that the coefficient of \(y'\) is nonzero at the point.
Use the point and computed slope in point-slope form, then verify that the line passes through the point.
When the Slope Formula Does Not Apply
In each derivation, isolating \(y'\) required division by a coefficient. For \(e^y+xy=1\), that coefficient is \(e^y+x\); for \(\ln(x+y)=y^2\), it is \(2y(x+y)-1\). If the relevant coefficient is zero, the displayed slope formula cannot be used at that point.
This limitation is specific: it says that this calculation does not determine \(y'\) by division. It does not automatically establish that the curve has no tangent, or that a local function branch cannot exist. A zero coefficient signals that the point requires separate analysis rather than substitution into the formula.
Check Your Understanding
For each question, keep track of the domain and of every factor arising from dependence on \(x\).
- Differentiate \(e^y+xy=1\) and solve for \(y'\), stating the condition required to divide.
- Differentiate \(\ln(x+y)=y^2\) and solve for \(y'\), stating both relevant conditions.
- Verify that \((1-e,1)\) lies on \(e^y+xy=1\), then find its slope.
- Find the tangent line to \(\ln(x+y)=y^2\) at \((1,0)\).
- What factor is missing in the incorrect derivative \(\frac{d}{dx}\ln(x+y)=1/(x+y)\), and where does it come from?
