Question

If \(m\) is the slope of a tangent to the curve \(e^y=1+x^2\) at \(x=1\), then \(m=\)

• A. \(\dfrac{2}{\log 2}\)
• B. \(\log 2\)
• C. \(2\)
• D. \(1\)

Hint:

In implicit differentiation, whenever \(e^y\) appears, remember: \[ \frac{d}{dx}(e^y)=e^y\frac{dy}{dx} \] using the chain rule.

Answer 1

The Correct Option is D

Step 1: Differentiate the given equation implicitly.
Given: \[ e^y=1+x^2 \] Differentiating both sides with respect to \(x\): \[ e^y\frac{dy}{dx}=2x \] Therefore, \[ \frac{dy}{dx}=\frac{2x}{e^y} \]

Step 2: Find \(y\) at \(x=1\).
Substitute \(x=1\) into the original equation: \[ e^y=1+1^2 \] \[ e^y=2 \]

Step 3: Find the slope at \(x=1\).
Using \[ \frac{dy}{dx}=\frac{2x}{e^y}, \] we get \[ m=\frac{2(1)}{2} \] \[ m=1 \]

Step 4: Final conclusion.
Hence, \[ \boxed{1} \]