Question
Question: If \(y=\sqrt{x}+\dfrac{1}{\sqrt{x}}\), prove that \(2x\dfrac{dy}{dx}+y=2\sqrt{x}\)....
If y=x+x1, prove that 2xdxdy+y=2x.
Solution
Hint: For solving this question we will first square the terms on the left-hand side and right-hand sides in the given equation. Then, we will try to get the equation xy2=x2+1+2x. After that, we will apply the product rule of differential calculus and formulas like dxd(xn)=nxn−1, dxd(yn)=(nyn−1)dxdy. Then, we will arrange the terms in the result to prove the desired result easily.
Complete step-by-step answer:
It is given that, y=x+x1 and we have to prove the following equation:
2xdxdy+y=2x
Now, before we proceed, we should know the following formula:
(a+b)2=a2+b2+2ab.............(1)
Now, we will square the terms on the left-hand side and right-hand sides in the given equation. Then,
y=x+x1⇒y2=(x+x1)2
Now, we will use the formula from the equation to write (x+x1)2=x+x1+2 in the above equation. Then,
y2=(x+x1)2⇒y2=(x)2+(x1)2+2×x×x1⇒y2=x+x1+2
Now, we will multiply the above equation by the variable x . Then,
y2=x+x1+2⇒xy2=x2+x×x1+2x⇒xy2=x2+1+2x...........................(2)
Now, we will use the formula from the equation (1) to write x2+1+2x=(x+1)2 in the above equation. Then,
xy2=x2+1+2x⇒xy2=(x+1)2
Now, we will take the square-root of the terms on the left-hand side and right-hand sides in the above equation. Then,
xy2=(x+1)2⇒xy2=(x+1)2⇒yx=(x+1)........................(3)
Now, before we proceed, we should know the following formulas and concepts of differential calculus:
1. If y=f(x)⋅g(x) , then dxdy=dxd(f(x)⋅g(x))=f′(x)⋅g(x)+f(x)⋅g′(x). This is also known as the product rule of differentiation.
2. If y=xn , then dxdy=dxd(xn)=nxn−1.
3. Derivative of yn with respect to x will be dxd(yn)=(nyn−1)dxdy .
Now, from the equation (2) we have the following equation:
xy2=x2+1+2x
Now, we will differentiate the above equation with respect to x . Then,
xy2=x2+1+2x⇒dxd(xy2)=dxd(x2+1+2x)
Now, we will use the product rule of differentiation to write dxd(xy2)=y2+2xydxdy in the above equation. Then,
dxd(xy2)=dxd(x2+1+2x)⇒dxd(x)×y2+x×dxd(y2)=dxd(x2)+dxd(1)+dxd(2x)
Now, we will use the formula dxd(xn)=nxn−1 to write dxd(x)=1 , dxd(x2)=2x , dxd(2x)=2 and formula dxd(yn)=(nyn−1)dxdy to write dxd(y2)=2ydxdy and as 1 is a constant term so, we can write dxd(1)=0 in the above equation. Then,
dxd(x)×y2+x×dxd(y2)=dxd(x2)+dxd(1)+dxd(2x)⇒y2+x×2ydxdy=2x+0+2⇒y2+2xydxdy=2(x+1)⇒y(y+2xdxdy)=2(x+1)⇒y+2xdxdy=y2(x+1)
Now, we will substitute (x+1)=yx from equation (3) in the above equation. Then,
y+2xdxdy=y2(x+1)⇒y+2xdxdy=y2yx⇒y+2xdxdy=2x⇒2xdxdy+y=2x
Now, from the above result, we conclude that, 2xdxdy+y=2x .
Hence proved.
Note: Here, the student should first understand what is asked in the question and then proceed in the right direction to prove the desired result quickly. And to avoid long calculations, we should first try to get an equation in x and y such that we will get a term like xdxdy after differentiating it. Moreover, we should apply the product rule of differential calculus, and formulas like dxd(xn)=nxn−1, dxd(yn)=(nyn−1)dxdy correctly without any mathematical error to prove the desired result easily.