MHT CET · Maths · Differentiation
If \(x^{2} y^{5}=(x+y)^{7}\), then \(\frac{d^{2} y}{d x^{2}}\) is equal to
- A \(y / x^{2}\)
- B \(x / y\)
- C 1
- D 0
Answer & Solution
Correct Answer
(D) 0
Step-by-step Solution
Detailed explanation
Given, \(x^{2} y^{5}=(x+y)^{7}\)
Taking log on both sides, we get
\(
2 \log x+5 \log y=7 \log (x+y)
\)
On differentiating, we get
\(\frac{2}{x}+\frac{5}{y} \frac{d y}{d x} =\frac{7}{x+y}\left(1+\frac{d y}{d x}\right) \)
\( \Rightarrow \frac{d y}{d x}\left(\frac{7}{x+y}-\frac{5}{y}\right)=\frac{2}{x}-\frac{7}{x+y}\)
\(
\Rightarrow \frac{d y}{d x}=\frac{y}{x}
\)
Again, differentiating, we get
\(\frac{d^{2} y}{d x^{2}} =\frac{x \frac{d y}{d x}-y}{x^{2}} \)
\( =\frac{x \cdot(y / x)-y}{x^{2}}[\text { from Eq. (i)] }\)
\( =0 \)
Taking log on both sides, we get
\(
2 \log x+5 \log y=7 \log (x+y)
\)
On differentiating, we get
\(\frac{2}{x}+\frac{5}{y} \frac{d y}{d x} =\frac{7}{x+y}\left(1+\frac{d y}{d x}\right) \)
\( \Rightarrow \frac{d y}{d x}\left(\frac{7}{x+y}-\frac{5}{y}\right)=\frac{2}{x}-\frac{7}{x+y}\)
\(
\Rightarrow \frac{d y}{d x}=\frac{y}{x}
\)
Again, differentiating, we get
\(\frac{d^{2} y}{d x^{2}} =\frac{x \frac{d y}{d x}-y}{x^{2}} \)
\( =\frac{x \cdot(y / x)-y}{x^{2}}[\text { from Eq. (i)] }\)
\( =0 \)
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