IV. Exercises: Find y', y", y'", and y"" for the following expressions: 1. y = x* 2. y = 2x2 + 3x - 5 3. y = x³(3x2 - 5) 4. y = 4x6 - 9x3 + 3x2 %3D

Q: 3(2x-1/2)=16

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Q: Simplify : (y-1 - (y + 2)-1) / 2

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Q: when y = In(2x – 4x). -

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Q: Find two First terms whose product is 3x. 3x? – 13xy + 4y² = (3x ?y)(x ?y)

A: Given that, 3x2-13xy+4y2=3x    ?yx   ?y

Q: Find : (a) 4xSy³ + 2x?y-y-2x-4 0. %3D

A: Basic identities of differentiation is used along with the chain rule

Q: x² + 4x - 4

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Q: 4/63 7x y

A: Simplify the given expression using the properties for exponents power rule. 3x6y34=3x6y314 fraction…

Q: Evaluate: f (3x2 - 20x)² dx

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Q: Write y=2(0.5x-6)(6-24x) in the form y=k(x-r)(x-s) and give the values of k,r,s.

A: We have to factor out 0.5 and -24

Q: Simplify: 512 + 5-3

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Q: X2 dx ex+3

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Q: In(x3 () Vx²_2X %3D domain

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Q: X=34-3 X=2-2y (-3,0) 12,07 y = 2X-4 -4 49-3

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Q: -dx 2x3/2 — Зх2

A: Given : ∫x2x32-3x2dx ∫x2x32-3x2dx = -∫x3x2-2x32dx Now, let us solve ∫x3x2-2x32dx Let us take  u=x⇒…

Q: Determine the points of inflection

A: According to power rule of differentiation .

Q: y = 1/2x + 5

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Q: (4x³y-$z+)(-2x-3 y7 z?) =D

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Q: -1/4+y=-2/5

A: We have been given a linear equation that we need to solve. In order to solve the given linear…

Q: d?y у-xV3 — 2х2; dx2

A: Here, the given function is: y=x3-2x2⇒y=3x2-2x4...(1)

Q: ID 2 = {(Xy) 2x=7

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Q: Differentiate y = (9x - 1)³. y = 8/9x - 9+

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Q: Simplify completely: (x7/2y1/2)3/2/x1/4y5/4

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Q: (24+2x-x2)1/2dx

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Q: Differentiate y- (92-)* g(x-x-4)² (+4-3)

A: Given query is to find the derivative of the function.

Q: Y + 7 = 4/3 (x -4)

A: We have to find the slope of the given line also graph the line in coordinate planes.

Q: y = In(3x² + 2x – 1)

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Q: Evaluate: 2x²-y y -z² by replacing x with 0.6, y with 0.32 and zwith 0.5.

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