find dy/dx if y =√x + 1/√x
Answers
Answer:
y=
x
+
x
1
y=x^{\frac{1}{2}}+x^{\frac{-1}{2}}y=x
2
1
+x
2
−1
\text{Differentiate with respect to x}Differentiate with respect to x
\displaystyle\frac{dy}{dx}=\frac{1}{2}x^{\frac{-1}{2}}-\frac{1}{2}x^{\frac{-3}{2}}
dx
dy
=
2
1
x
2
−1
−
2
1
x
2
−3
\displaystyle\frac{dy}{dx}=\frac{1}{2\sqrt{x}}-\frac{1}{2\,x\sqrt{x}}
dx
dy
=
2
x
1
−
2x
x
1
\text{Now,}Now,
\displaystyle\,2x\,\frac{dy}{dx}+y2x
dx
dy
+y
=\displaystyle,2x(\frac{1}{2\sqrt{x}}-\frac{1}{2\,x\sqrt{x}})+\sqrt{x}+\frac{1}{\sqrt{x}}=,2x(
2
x
1
−
2x
x
1
)+
x
+
x
1
=\displaystyle\frac{2x}{2\sqrt{x}}-\frac{2x}{2\,x\sqrt{x}}+\sqrt{x}+\frac{1}{\sqrt{x}}=
2
x
2x
−
2x
x
2x
+
x
+
x
1
=\displaystyle\sqrt{x}-\frac{1}{\sqrt{x}}+\sqrt{x}+\frac{1}{\sqrt{x}}=
x
−
x
1
+
x
+
x
1
=\displaystyle\,2\,\sqrt{x}=2
x
\implies\boxed{\bf\,2x\,\frac{dy}{dx}+y=2\,\sqrt{x}}⟹
2x
dx
dy
+y=2
x