Question

Differentiating \[(1-x^2)y_{1}^{2}=1\] gives which expression?

पर्याय

• \[(1-x^2)\cdot y_{1}y_{2}+y_{1}^{2}(0-2x)=1\]

• \[(1-x^2)\cdot2y_{1}y_{2}+y_{1}^{2}(0-2x)=0\]

• \[(1-x^2)\cdot2y_{1}y_{2}-y_{1}^{2}(0-2x)=0\]

• \[(1-x^2)\cdot2y_{1}^{2}+y_{2}(0-2x)=0\]

Answer 1

Apply the product rule to \[(1-x^2)y_{1}^{2}\]. The derivative of \[y_{1}^{2}\] is \[2y_{1}y_{2}\], and the derivative of \[1\] is \[0\].