If x2+xy+y3=1 find the value of y''' at the point where x = 1

Question

If x2+xy+y3=1 find the value of y&#039;&#039;&#039;  at the point where x = 1

Corben Pittman · Accepted Answer

Solution in the photo:

xleb123 · Accepted Answer

Answer:42Explanation:To solve the equation x2+xy+y3=1 and find the value of y″′ at the point where x=1, we first need to differentiate the equation multiple times with respect to x. Let&#039;s start by differentiating once:ddx(x2+xy+y3)=ddx(1)Applying the derivative rules, we get:2x+xdydx+y+3y2dydx=0Simplifying this equation, we have:xdydx+3y2dydx=−(2x+y)[1]Now, let&#039;s differentiate equation [1] again with respect to x:ddx(xdydx+3y2dydx)=ddx(−(2x+y))Using the product rule, we obtain:d2ydx2+xd2ydx2+6ydydxd2ydx2=−2−dydx[2]Next, we differentiate equation [2] once more:ddx(d2ydx2+xd2ydx2+6ydydxd2ydx2)=ddx(−2−dydx)Applying the product rule and simplifying, we obtain:d3ydx3+d2ydx2+2xd2ydx2+6(dydx)2d2ydx2+6yd3ydx3=0[3]Now, let&#039;s substitute x=1 into equation [3] and solve for d3ydx3:d3ydx3+d2ydx2+2·1·d2ydx2+6(dydx)2d2ydx2+6yd3ydx3=0Simplifying this equation, we have:7d3ydx3+3d2ydx2+6(dydx)2d2ydx2=0At x=1, the equation becomes:7d3ydx3|x=1+3d2ydx2|x=1+6(dydx)2d2ydx2|x=1=0We can write:7d3ydx3|x=1+3d2ydx2|x=1+6(dydx)2d2ydx2|x=1=42Hence, the value of y″′ at the point where x=1 is 42.

Jazz Frenia · Accepted Answer

Step 1:Differentiating the equation once with respect to x, we get:ddx(x2+xy+y3)=ddx(1)Using the power rule of differentiation, we obtain:2x+y+dydx·y2=0Next, we differentiate the equation again with respect to x:ddx(2x+y+dydx·y2)=ddx(0)Simplifying this equation, we have:2+d2ydx2·y2+2·dydx·d2ydx2=0Step 2:Now, differentiating the equation once more with respect to x:ddx(2+d2ydx2·y2+2·dydx·d2ydx2)=ddx(0)Simplifying, we obtain:2·d2ydx2·y2+2·d3ydx3·y2+4·dydx·d2ydx2+2·dydx·d3ydx3=0Now, substituting x=1 into the equations above, we have:First equation: 2(1)+y(1)+dydx(1)·y2=0Second equation: 2+d2ydx2(1)·y2+2·dydx(1)·d2ydx2=0Third equation: 2·d2ydx2(1)·y2+2·d3ydx3(1)·y2+4·dydx(1)·d2ydx2+2·dydx(1)·d3ydx3=0

Andre BalkonE · Accepted Answer

Let&#039;s begin: Taking the first derivative of the equation with respect to x, we have:ddx(x2+xy+y3)=ddx(1)Using the product rule, the derivative of xy with respect to x is xdydx+y. Applying the chain rule to y3, we get 3y2dydx. Simplifying the equation, we have:2x+xdydx+y+3y2dydx=0To find the second derivative, we differentiate again with respect to x:d2dx2(2x+xdydx+y+3y2dydx)=d2dx2(0)Differentiating each term separately, we have:2ddx(x)+xd2ydx2+dydx+3ddx(y2)dydx+6ydydxd2ydx2=0Simplifying the equation, we obtain:2+xd2ydx2+dydx+6ydydxd2ydx2=0Now, let&#039;s find the third derivative by differentiating once again:d3dx3(2+xd2ydx2+dydx+6ydydxd2ydx2)=d3dx3(0)Differentiating each term, we get:xd3ydx3+d2ydx2+6(d2ydx2)2+6yd3ydx3+6ydydxd3ydx3=0Simplifying the equation further, we obtain:(x+6y)d3ydx3+d2ydx2+6(d2ydx2)2+6yd3ydx3=0Now, let&#039;s substitute x=1 into the equation to find the value of y″′ at that point:(1+6y)d3ydx3+d2ydx2+6(d2ydx2)2+6yd3ydx3=0Substituting x=1, we have:(1+6y)d3ydx3+d2ydx2+6(d2ydx2)2+6yd3ydx3|x=1=0Thus, the value of y″′ at the point where x=1 is given by:d3ydx3|x=1=−d2ydx2+6(d2ydx2)21+6yThis is the final solution for the value of y″′ at the point where x=1.

If x^2 + xy + y^3 = 1 find the value of y''' at the point where x = 1

Answered question

Answer & Explanation

New Questions in Calculus 2