Question

If $3x^2-4xy=1$, then when $x=1,\frac{dy}{dx}=$

• A. $\frac{3}{2}$
• B. $1$
• C. $\frac{1}{2}$
• D. $0$
• E. $-\frac{1}{2}$

Answer 1

Answer: (B)

$1$

Given $3x^2-4xy=1$
Differentiate with respective to $x$ on both sides, we get

$6x-4(y+x\frac{dy}{dx})=0$
$\Rightarrow \frac{1}{x}(\frac{3x}{2}-y)=\frac{dy}{dx}$
Given $x=1$, so we get $3-4y=1$

$\Rightarrow y=\frac{1}{2}$
Therefore $\frac{dy}{dx}=1(\frac{3}{2}-\frac{1}{2})=1\left(1\right)=1$