Question 1: Evaluate all the following
i) limx→∞8x+12x+3
Step 1: Divide the numerator and denominator by the highest power of x, which is x.
limx→∞x8x+x1x2x+x3=limx→∞8+x12+x3
Step 2: Apply the limit. As x→∞, x3→0 and x1→0.
8+02+0=82
Step 3: Simplify the fraction.
82=41
The limit is 41.
ii) limx→0x2+22x2+8x+3
Step 1: Substitute x=0 into the expression, as the denominator is not zero at x=0.
(0)2+22(0)2+8(0)+3
Step 2: Simplify the expression.
0+20+0+3=23
The limit is 23.
Question 2: Find dxdy of the following
i) y=4x+33x2−8
Step 1: Apply the quotient rule, dxdy=v2u′v−uv′, where u=3x2−8 and v=4x+3.
Then u′=6x and v′=4.
dxdy=(4x+3)2(6x)(4x+3)−(3x2−8)(4)
Step 2: Expand the terms in the numerator.
dxdy=(4x+3)224x2+18x−(12x2−32)
Step 3: Simplify the numerator by distributing the negative sign and combining like terms.
dxdy=(4x+3)224x2+18x−12x2+32=(4x+3)212x2+18x+32
The derivative is (4x+3)212x2+18x+32.
ii) y=(3x3−8x2−1)6
Step 1: Apply the chain rule, dxdy=n[f(x)]n−1⋅f′(x), where f(x)=3x3−8x2−1 and n=6.
First, find f′(x): f′(x)=dxd(3x3−8x2−1)=9x2−16x.
dxdy=6(3x3−8x2−1)6−1⋅(9x2−16x)
Step 2: Simplify the exponent and rearrange the terms.
dxdy=6(9x2−16x)(3x3−8x2−1)5
The derivative is 6(9x2−16x)(3x3−8x2−1)5.
iii) 4xy+y2=x2+2
Step 1: Differentiate both sides of the equation with respect to x. Remember to use the product rule for 4xy and the chain rule for y2.
dxd(4xy)+dxd(y2)=dxd(x2)+dxd(2)
(4⋅y+4x⋅dxdy)+(2y⋅dxdy)=2x+0
4y+4xdxdy+2ydxdy=2x
Step 2: Group terms containing dxdy on one side and other terms on the other side.
4xdxdy+2ydxdy=2x−4y
Step 3: Factor out dxdy from the terms on the left side.
dxdy(4x+2y)=2x−4y
Step 4: Solve for dxdy by dividing both sides by (4x+2y).
dxdy=4x+2y2x−4y
Step 5: Simplify the expression by dividing the numerator and denominator by 2.
dxdy=2(2x+y)2(x−2y)=2x+yx−2y
The derivative is 2x+yx−2y.
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