Resourcesdown chevron
Testimonials
Plans
Sign in
Sign up
Resourcesright arrow
Testimonials
Plans
Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

0=-(y^(2)-2y+x)

0=−(y2−2y+x) 0=-\left(y^{2}-2 y+x\right)

View step-by-step helpView step-by-step help
Homeright chevron icon
Math Problemsright chevron icon
Calculusright chevron icon
Find derivatives using the chain rule Iright chevron icon

Full solution

Q. 0=−(y2−2y+x) 0=-\left(y^{2}-2 y+x\right)
  1. Identify equation: Identify the equation to be solved. The equation is 0=−(y2−2y+x)0=-(y^{2}-2y+x).
  2. Rewrite without negative sign: Rewrite the equation without the negative sign by multiplying both sides by −1-1 to get a standard quadratic form. The equation becomes y2−2y+x=0y^{2}-2y+x=0.
  3. Recognize quadratic form: Recognize that the equation is a quadratic equation in the form of ay2+by+c=0ay^2 + by + c = 0, where a=1a=1, b=−2b=-2, and c=xc=x.
  4. Use quadratic formula: To solve the quadratic equation, we can either factor it, complete the square, or use the quadratic formula. Since we do not have specific values for yy and xx, we will use the quadratic formula: y=−b±b2−4ac2ay = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
  5. Substitute values into formula: Substitute the values of aa, bb, and cc into the quadratic formula. This gives us y=−(−2)±(−2)2−4⋅1⋅x2⋅1.y = \frac{-(-2) \pm \sqrt{(-2)^2 - 4\cdot1\cdot x}}{2\cdot1}.
  6. Simplify equation: Simplify the equation by performing the operations inside the formula. This gives us y=2±4−4x2y = \frac{2 \pm \sqrt{4 - 4x}}{2}.
  7. Further simplify equation: Further simplify the equation by dividing the terms inside the square root as well as the 22 outside. This gives us y=1±1−xy = 1 \pm \sqrt{1 - x}.

More problems from Find derivatives using the chain rule I

Question
Find the derivative of f(x) f(x) .\newlinef(x)=cos⁡(x) f(x) = \cos(x) \newlinef′(x)= f'(x) = ______
Get tutor helpright-arrow

Posted 1 month ago

Question
Find the derivative of f(x) f(x) .\newlinef(x)=tan⁡(x) f(x) = \tan(x) \newlinef′(x)= f'(x) = ______
Get tutor helpright-arrow

Posted 2 months ago

Question
Find the derivative of f(x) f(x) .\newlinef(x)=ex f(x) = e^x \newlinef′(x)= f'\left(x\right) = ______
Get tutor helpright-arrow

Posted 2 months ago

Question
Find the derivative of f(x) f(x) .\newlinef(x)=ln⁡(x) f(x) = \ln(x) \newlinef′(x)= f'(x) = ______
Get tutor helpright-arrow

Posted 2 months ago

Question
Find the derivative of f(x) f(x) .\newlinef(x)=tan⁡−1x f(x) = \tan^{-1}{x} \newlinef′(x)= f'(x) = ______
Get tutor helpright-arrow

Posted 2 months ago

Question
Find the derivative of f(x) f(x) .\newlinef(x)=xex f(x) = x e^x \newlinef′(x)= f'\left(x\right) = ______
Get tutor helpright-arrow

Posted 2 months ago

Question
Find the derivative of f(x) f(x) . \newlinef(x)=exx f(x) = \frac{e^x}{x} \newlinef′(x)= f'(x) = ______
Get tutor helpright-arrow

Posted 2 months ago

Question
Find the derivative of f(x) f(x) . \newline f(x)=e(x+1) f(x) = e^{(x + 1)} \newline f′(x)= f'\left(x\right) = ______
Get tutor helpright-arrow

Posted 1 month ago

Question
Find the derivative of f(x) f(x) . \newlinef(x)=x+3 f(x) = \sqrt{x+3} \newlinef′(x)= f'(x) = ______
Get tutor helpright-arrow

Posted 2 months ago

Question
Find the derivative of g(x) g(x). \newline g(x)=ln⁡(2x) g(x) = \ln(2x) \newline g′(x)= g^{\prime}(x) = ______
Get tutor helpright-arrow

Posted 3 months ago

View all (70)