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: Take ln of both sides to make y explicit.
$ e^y=1+x^2 \implies y=\ln(1+x^2) $.
Step 2: Differentiate using the chain rule.
\[ \frac{dy}{dx} = \frac{2x}{1+x^2} \]
Step 3: Find the point at $ x=1 $.
$ y=\ln(1+1)=\ln 2 $. Point is $ (1,\ln 2) $.
Step 4: Evaluate the slope at $ x=1 $.
\[ m = \frac{2\cdot 1}{1+1} = \frac{2}{2} = 1 \]
Step 5: Write the tangent line.
$ y-\ln 2=1\cdot(x-1) \implies y=x-1+\ln 2 $. The slope $ =1 $ means the tangent makes a $ 45^\circ $ angle with the x-axis.
Step 6: State the slope.
\[ \boxed{1} \]