WBJEE · Maths · Differentiation
Let \(y=\left(\frac{3^{x}-1}{3^{x}+1}\right) \sin x+\log _{e}(1+x) \quad x>-1\) Then, at \(x=0, \frac{d y}{d x}\) equals
- A 1
- B 0
- C -1
- D -2
Answer & Solution
Correct Answer
(A) 1
Step-by-step Solution
Detailed explanation
Given, \(y=\left(\frac{3^{x}-1}{3^{x}+1}\right) \sin x+\log _{e}(1+x), x>-1\) Differentiating w.r.t. \(x\), we get \(\frac{d y}{d x}=\frac{d}{d x}\left[\left(\frac{3^{x}-1}{3^{x}+1}\right) \sin x\right]+\frac{d}{d x} \log _{e}(1+x)\)…
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