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a curve has an equation: y = x^2 - 2x - 24x^0.5 x>0 find dy/dx and d^2y/dx^2

dy/dx = 2x -2 - 12x^-0.5d^2/dx^2 = 2 + 6 x^-3/2
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Solve (x-1)^2-2(x-1)-3=0

Solving this equation would mean finding the values of x that satisfy it equating to zero. Therefore, to start, you must simplify the equation by expanding all the values to get (x^2-2x+1)-(2x-2)-3=0 [expand...
EL
Answered by Emily L. • Maths tutor
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Solve the equation x^6 + 26x^3 − 27 = 0

While this equation may look complicated, it's actually much easier than it looks - this equation is called a hidden quadratic. This is because it can be rewritten and solved the same way you would solve a q...
SP
Answered by Saachi P. • Maths tutor
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Find the equation of the normal to the curve 2x^3+3xy+2/y=0 at the point (1,-1)

Step 1 Use Implicit differentiation with respect to x and y - 6x^2+3y + 3x(dy/dx) - 2/y^2(dy/dx) = 0Step 2 Write the equation as dy/dx =... - dy/dx = (6x^2 + 3y)/(2/y^2 - 3x)Step 3 Input (1,-1) into the equa...
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Solve the following equation: 13y - 5 = 9y + 27

13y - 5 = 9y + 2713y - 9y = 27 + 54y = 32y = 32/4 y = 8
MG
Answered by Megan G. • Maths tutor
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