MHT CET · Maths · Differentiation
If \(\mathrm{f}(x)=(1+x)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\), then \(f^{\prime}(1)=\)
- A 60
- B 80
- C 240
- D 120
Answer & Solution
Correct Answer
(D) 120
Step-by-step Solution
Detailed explanation
\(y=(1+x)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)...(i)\)
Taking 'log' on both sides, we get
\(\begin{aligned}
& \log y=\log (1+x)+\log \left(1+x^2\right)+\log \left(1+x^4\right) \\
&+ \log \left(1+x^8\right)
\end{aligned}\)
Differentiating w.r.t. \(x\), we get
\(\frac{1 \mathrm{~d} y}{y}=\frac{1}{1+x}+\frac{2 x}{1+x^2}+\frac{4 x^3}{1+x^4}+\frac{8 x^7}{1+x^8}...(ii)\)
At \(x=1\), (i) \(\Rightarrow y=16\)
\(\therefore \quad\) (ii) \(\Rightarrow\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=1}=16\left(\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2}\right)=120\)
Taking 'log' on both sides, we get
\(\begin{aligned}
& \log y=\log (1+x)+\log \left(1+x^2\right)+\log \left(1+x^4\right) \\
&+ \log \left(1+x^8\right)
\end{aligned}\)
Differentiating w.r.t. \(x\), we get
\(\frac{1 \mathrm{~d} y}{y}=\frac{1}{1+x}+\frac{2 x}{1+x^2}+\frac{4 x^3}{1+x^4}+\frac{8 x^7}{1+x^8}...(ii)\)
At \(x=1\), (i) \(\Rightarrow y=16\)
\(\therefore \quad\) (ii) \(\Rightarrow\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=1}=16\left(\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2}\right)=120\)
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