Differentiation

Find the gradient of the curve at the point where the curve `y=frac{5x-10}{x^2}` crosses the x-axis.

The curve crosses the x-axis where `y=o`

`y=frac{5x-10}{x^2}` becomes `0=frac{5x-10}{x^2}`

`5x-10=0`

`x=2`

Rewrite `y=frac{5x-10}{x^2}`

`y=frac{5x}{x^2} - frac{10}{x^2}= 5x^-1 - 10x^-2`

`frac{dy}{dx} = -5x^-2 + 20x^-3 =frac{-5}{x^2} + frac{20}{x^3}`

 Substitute `x=2` into `frac{dy}{dx}`

`frac{dy}{dx}=frac{-5}{2^2}+frac{20}{2^3}=frac{-5}{4}+frac{20}{8}=frac{5}{4}`

Then the gradient is `frac{5}{4}`

Given that `y=2x^3 - 3x^2-36x+5`, find the range of values of `x` for which `frac{dy}{dx} < 0` .

Differentiate`y=`-2`2x^3-3x^2-36x+5`

`frac{dy}{dx}= 6x^2-6x-36`

`6x^2-6x-36 <0`(as `frac{dy}{dx}<0`)

 `x^2-x-6<0`

`(x-3)(x+2)<0`

 The graph of `y=(x-3)(x+2)` is a `uu` shaped parabola

The x-intercepts are at `x=3` and `x=-2`

The range of values of `x` for which the curve is negative (below the x-axis)

Then `-2ltxlt3`

Differentiate with respect to `x`:

`a. frac{7}{(2x^2-5x)^7}`

`b. frac{6}{root(3)(2-3x)}`

`a.`Let `y= frac{7}{(2x^2 - 5x)^7`

Rewrite as `y= 7(2x^2 - 5x)^-7`

Let `u=2x^2-5x` , then`y=7u^-7`

`frac{du}{dx}=4x-5`

`frac{dy}{du}=-49u^-8`

 Use the chain rule to differentiate

`frac{dy}{dx}=frac{dy}{du} xx frac{du}{dx}`

`=-49u^-8 xx(4x-5)`

`=-frac{49(4x-5)}{u^8}`

`=-frac{49(4x-5)}{(2x^2-5x)^8`

`=-frac{49(4x-5)}{[x(2x-5)]^8`

`=-frac{49(4x-5)}{x^8(2x-5)^8`

`b.`Let `y=frac{6}{root(3)(2-3x)}`

Rewrite as`y= 6(2-3x)^(-1/3)`

Let `u=2-3x` , then`y=6u^(-1/3)`

`frac{du}{dx}=-3`

`frac{dy}{du}=-1/3 xx6u^(-4/3) = -2u^(-4/3)`

 Use the chain rule to differentiate

`frac{dy}{dx}=frac{dy}{du}xxfrac{du}{dx}`

`=-2u^(-4/3) xx(-3)`

`=6/u^(4/3)`

`=frac{6}{(2-3x)^(4/3)`

`=frac{6}{root(3)((2-3x)^4)}`

Find the coordinates of the point on the curve `y=sqrt ((x^2-10x+26))` where the gradient is `0`.

Rewrite `y= sqrt((x^2-10x+26))` as `y=(x^2-10x+26)^(1/2)`

Let `u=x^2-10x+26` , then `y=u^(1/2)`

`frac{du}{dx}=2x-10`

`frac{dy}{du}=1/2 u^(-1/2)`

Use the chain rule to differentiate

`frac{dy}{dx}=frac{dy}{du}xxfrac{du}{dx}`

`=1/2u^(-1/2)xx(2x-10)`

`=frac{1}{2(x^2-10x+26)^(-1/2)} xx(2x-10)`

`=frac{x-5}{sqrt(x^2-10x+26}`

The gradient is zero when `frac{x-5}{sqrt(x^2-10x+26)}=0`

`x-5=0`

`x=5`

Substitute `x=5` into `y=sqrt((x^2-10x+26))`

`y=sqrt((5^2-10(5)+26))=1`

The coordinates of the point are `(5,1)`

The normal to the curve `y=x^3 -5x+3` at the point `(-1,7)` intersects the y-axis at the point P. Find the coordinates of P.

Differentiate `y=x^3-5x+3`

`frac{dy}{dx}=3x^2 -5`

The gradient of the curve at `x=-1`

 `3(-1)^2-5=-2`

The curve `y= 2x^2 +kx -3` at the point `(3,-6)` is parallel to the line `x+5y=10` .

a. Find the value of `k`.

b. Find the coordinates of the point where the normal meets the curve again.

a.Differentiate `y=2x^2 +kx -3`

`frac{dy}{dx}=4x+k`

The gradient of the curve at `x=3`

`4(3) +k =12+k`

The gradient of the normal `=-frac{1}{12+k}`  (as `m_1xxm_2=-1` )

The gradient of the line `x+5y=10`  is found by rearranging this equation and comparing it with

`y=mx+c`

`5y=-x+10`

`y=-1/5x +2`

The gradient `m=-1/5`

Then `-1/5 =-frac{1}{12+k}`

`12+k=5`

`k=-7`

b.Substitute `k=-7`  into `m=- frac{1}{12+k}`

`-frac{1}{12+(-7)}=-1/5`

The equation of the normal

`y-y_1=m(x-x_1)`

`y-(-6)=-1/5(x-3)`(as `m=-1/5` , `x_1=3` , `y_1=-6`)

`y+6=-1/5x+3/5`

`y=-1/5x-27/5` (1)

Substitute `k=-7`  into `y=2x^2+kx-3`

`y=2x^2-7x-3` (2)

From (1) and (2)

`2x^2-7x-3=-1/5x-27/5`

`10x^2-35x-15=-x-27`

`10x^2-34x+12=0`

Using quadratic formula `x=frac{-b+-sqrt(b^2-4ac)}{2a}`

`x=frac{-(-34)+-sqrt((-34)^2-4(10)(12))}{2(10)`

`x=frac{34+-sqrt(676)}{20}`

`x=0.4` or `x=3`

Then `x=0.4`

Substitute `x=0.4` into (1)

`y=-1/5(0.4)-27/5=-5.48`

The point where the normal meets the curve again has coordinates `(0.4, −5.48)`

Given that `f(x)= frac{2}{sqrt(1-2x)}` , find the value of `f''(-4)`

Rewrite `f(x) frac{2}{sqrt(1-2x)}`  as `f(x)=2(1-2x)^(-1/2)`

Use the chain rule to differentiate twice

`f'(x)=-1(1-2x)^(-3/2) xx(-2)=2(1-2x)^(-3/2)`

`f''(x)=-3(1-2x)^(-5/2)xx(-2)=6(1-2x)^(-5/2)`

`f''(-4)=6[1-2(-4)]^(-5/2)=6xx9^(-5/2)=6/sqrt(9)^5=2/81`

Given that `y=x^2-2x+5` , show that `4frac{d^2y}{dx^2}+(x-1)frac{dy}{dx}=2y`

Differentiate `y=x^2-2x+5`

`frac{dy}{dx}=2x-2`

`frac{d^2y}{dx^2}=2`

Substitute into `4frac{d^2y}{dx^2}+(x-1)frac{dy}{dx}`

`4xx2+(x-1)(2x-2)`

`=8+2x^2-2x-2x+2`

`=2x^2-4x+10`

As`2y=2(x^2-2x+5)=2x^2-4x+10`

Then `4frac{d^2y}{dx^2}+(x-1)frac{dy}{dx} =2y`

A curve has equation `y=x^3+2x^2-4x+6` .

a. Show that `frac{dy}{dx}=0` when `x=-2` and when `x=2/3` .

b. Find the value of `frac{d^2y}{dx^2}` when `x=-2` and when `x=2/3` .

Differentiate `y=x^3 +2x^2-4x+6`

`frac{dy}{dx}=3x^2+4x-4`(1)

`frac{d^2y}{dx^2}=6x+4`(2)

a. Substitute `x=-2` into (1)

`frac{dy}{dx}=3(-2)^2+4(-2)-4`

`frac{dy}{dx}=0`

 Substitute `x=2/3` into (1)

 `frac{dy}{dx}=3(2/3)^2+4(2/3)-4`

`frac{dy}{dx}=0`

 b. Substitute `x=-2` into (2)

 `frac{d^2y}{dx^2}=6xx(-2)+4`

`frac{d^2y}{dx^2}=-8`

Substitute `x=2/3` into (2)

`frac{d^2y}{dx^2}=6xx2/3+4`

`frac{d^2y}{dx^2}=8`

A curve has equation `y=frac{ax+b}{x^2}` . Given that `frac{ dy}{dx}=0` and `frac{d^2y}{dx^2}=1/2` when `x=2` , find the value of a and the value of `b`. 

Rewrite `y=frac{ax+b}{x^2}` as `y=frac{ax}{x^2}+frac{b}{x^2}=ax^-1+bx^-2`

Differentiate `y=ax^-1+bx^-2`

`frac{dy}{dx}=-ax^-2-2bx^-3` (1)

`frac{d^2y}{dx^2}=2ax^-3+6bx^-4` (2)

Substitute `x=2` into (1)

 `frac{dy}{dx}=-a(2)^-2-2b(2)^-3`

`frac{dy}{dx}=-frac{a}{4}-frac{2b}{8}`

`frac{dy}{dx}=frac{-a-b}{4}` 

`0=frac{-a-b}{a}` (as `frac{dy}{dx}=0`)

`-a-b=0`

`b=-a` (3)

Substitute `x=2` into (2)

`frac{d^2y}{dx^2}=2a(2)^-3+6b(2)^-4`

`frac{d^2y}{dx^2}=(2a)/8 +(6b)/16`

`frac{d^2y}{dx^2}=frac{2a+3b}{8}`

`1/2=frac{2a+3b}{8}` (as `frac{d^2y}{dx^2}=1/2` )

`8=2(2a+3b)`

`2a+3b=4`(4)

From (3) and (4)

`2a+3(-a)=4`

`a=-4`

Substitute `a=-4` into (3)

 `b=-(-4)=4`

The diagram shows a solid cone which has a slant height of `15` cm and a vertical height `h` cm.

a. Show that the volume `V` cm³, of the cone is given by `v=1/3 pi(225h-h^3)`.

[The volume of a cone of radius `r` and vertical height  `h`is `1/3 pi r^2h` ]

b. Given that `h` can vary, find the value of `h` for which `V` has a stationary value. Determine, showing all necessary working, the nature of this stationary value.

a. Let `r` is the radius of the cone

Use Pythagoras

`r^2+h^2=15^2`

`r^2=225-h^2`

For `V=1/3pir^2h`

`v=1/3pi(225-h^2)h`

`v=1/3pi(225h-h^3)`

b. Differentiate `V=1/3pi(225h-h^3)`

`frac{dv}{dh}=1/3pi(225-3h^2)`

`frac{d^2v}{dh^2}=1/3pi(0-6h)=-2pih` (1)

For `V` has a stationary value

`1/3pi(225-3h^2)=0` (as `frac{dv}{dh}=0`)

`1/3pi(225-3h^2)=0`

`225-3h^2=0`

`3h^2=225`

`h^2=75`

`h=5sqrt3`

Substitute `h=5sqrt3` into (1)

`frac{d^2v}{dh^2}=-2pi(5sqrt3)=-10pisqrt3`

`frac{d^2v}{dh^2}<0`

Then `V` is a maximum value

The normal to the curve `y = x^3 - 5x + 3` at the point `(-1, 7)` intersects the y-axis at the point P.

Find the coordinates of `P`.

The gradient of the normal `=1/2` (as `m_1xxm_2=-1`)

Equation of the normal

 `y-y_1=m(x-x_1)`

`y-7=1/2[x-(-1)]` (as `m=1/2`, `x_1=-1`, `y_1=7`)

 `y-7=1/2x+1/2`

 `2y-14=x+1`

`2y=x+15`

 `2y=0+15` (`x=0` as the normal intersects y-axis at P)

 `y=7.5`

Then P has coordinates `(0, 7.5)`

A curve is such that `frac{dy}{dx}=x^3-frac{4}{x^2}`. The point `P (2, 9)` lies on the curve. A point moves on the curve in such a way that the x-coordinate is decreasing at a constant rate of `0.05` units per second. Find the rate of change of the y-coordinate when the point is at P.

Let `frac{dx}{dt}=-0.05`

`frac{dy}{dx}=x^3-frac{4}{x^2}` at the point `P (2, 9)`

 `(2)^3-frac{4}{(2)^2}=8-1=7`

Use the chain rule 

 `frac{dy}{dt}=frac{dy}{dx}xxfrac{dx}{dt}`

 `frac{dy}{dt}=7xx(-0.05)`

 `frac{dy}{dt}=-0.35`

Then the y-coordinate is decreasing at a rate of `0.35` units per second

A curve has equation `y=(2x-1)^-1 +2x` .

a. Find `frac{dy}{dx}` and `frac{d^2y}{dx^2}`

b. Find the x-coordinates of the stationary points and, showing all necessary working, determine the nature of each stationary point.

`a.`Differentiate `y=(2x-1)^-1+2x`

`frac{dy}{dx}=-2(2x-1)^-2+2`

`frac{d^2y}{dx^2}=8(2x-1)^-3` (1)

`b.`For the curve has a stationary value

`-2(2x-1)^-2+2=0` (as `frac{dy}{dx}=0`)

`-2(2x-1)^-2=-2`

`frac{1}{(2x-1)^2}=1`

 `2x-1=1`or`2x-1=-1`

`x=1` or `x=0`

Substitute `x=1`  into (1)

`8[2(1)-1]^-3=8`

`frac{d^2y}{dx^2}>0` then `x=1` is a minimum value 

Substitute `x=0` into (1)

`8[2(0)-1]^-3=-8`

`frac{d^2y}{dx^2}<0` then `x=0` is a maximum value

The equation of a curve is `y=2x+1+frac{1}{2x+1}` for `x > -1/2` .

a. Find `(dy)/(dx)` and `(d^2y)/(dx^2)`

b. Find the coordinates of the stationary point and determine the nature of the stationary point.

a. Rewrite `y=2x+1+frac{1}{2x+1}` as `y=2x+1+(2x+1)^-1`

Differentiate `y=2x+1+(2x+1)^-1`

`(dy)/(dx)=2-2(2x+1)^-2=2-frac{2}{(2x+1)^2`

`frac{d^2y}{dx^2}=-2(-4)(2x+1)^-3=frac{8}{(2x+1)^3`

b. For the curve has a stationary value

`2-frac{2}{(2x+1)^2}=0`(as `(dy)/(dx)=0` )

 `frac{2}{(2x+1)^2}=2`

 `frac{1}{(2x+1)^2}=1`

`2x+1=1`or`2x+1=-1`

`x=0` or `x=-1`

Then `x=0` (as `x> -1/2`)

Substitute `x=0` into `y=2x+1+frac{1}{2x+1}`

 `y=2(0)+1+frac{1}{2(0)+1}=2`

The stationary point has coordinates `(0, 2)`

At `(0, 2)`

 `frac{d^2y}{dx^2}=frac{8}{[2(0)+1]^3}=8`

`frac{d^2y}{dx^2}>0` then `(0, 2)` is a minimum point

Air is being pumped into a balloon in the shape of a sphere so that its volume is increasing at a constant rate of `50` cm³s^-1.

Find the rate at which the radius of the balloon is increasing when the radius is `10` cm.

The volume `V=4/3pir^3`

Differentiate `V=4/3pir^3`

 `frac{dv}{dr}=4/3pi(3r^2)=4pir^2`

Let `r=10`

 `frac{dv}{dr}=4pi(10)^2=400pi`

 `frac{dr}{dv}=frac{1}{(dv)/(dr)}=frac{1}{400pi}`

Let `frac{dv}{dt}=50`

Use the chain rule 

 `frac{d}{dt}=frac{dr}{dv}xxfrac{dv}{dt}`

 `frac{dr}{dt}=frac{1}{400pi}xx50`

 `frac{dr}{dt}=frac{1}{8pi}`

Then the rate is `frac{1}{8pi}` cm/s

The equation of a curve is `y=2+sqrt(25-x^2)` .

Find the coordinates of the point on the curve at which the gradient is `4/3` .

Rewrite  `y=2+sqrt(25-x^2)`as `y=2+(25-x^2)^(1/2)`

Differentiate `y=2+(25-x^2)^(1/2)`

`frac{dy}{dx}=1/2(-2x)(25-x^2)^(-1/2)=-frac{x}{sqrt(25-x^2)`

`4/3=-frac{x}{sqrt(25-x^2)` (as `frac{dy}{dx}=4/3`)

`16/9=frac{x^2}{25-x^2}`

`16(25-x^2)=9x^2`

`400-16x^2=9x2`

`400=25x^2`

`x^2=16`

`x=4` or `x=-4`

Substitute `x=4` into `(dy)/(dx)=-frac{x}{sqrt(25-x^2)`

`-frac{4}{sqrt(25-(4)^2)}=-4/3ne4/3`

Then `x=-4`

Substitute `x=-4` into `y=2+sqrt(25-x^2)`

`2+sqrt(25-(-4)^2)=5`

The point has coordinates `(-4, 5)`

The volume `V` m³ of a large circular mound of iron ore of radius `r` m is modelled by the equation `V=3/2(r-1/2)^3-1` for `rge2` . Iron ore is added to the mound at a constant rate of `1.5` m³ per second.

a. Find the rate at which the radius of the mound is increasing at the instant when the radius is `5.5` m.

b. Find the volume of the mound at the instant when the radius is increasing at `0.1` m per second.

a. Differentiate `V=3/2(-1/2)^3-1`

`(dv)/(dr)=3/2(3)(r-1/2)^2=9/2(r-1/2)^2` (1)

Let `r=5.5`

 `(dv)/(dr)=9/2(5.5-1/2)^2=112.5`

 `(dr)/(dr)=frac{1}{(dv)/(dr)}=1/112.5`

Let `(dv)/(dt)=1.5`

Use the chain rule 

 `(dr)/(dt)=(dr)/(dv)xx(dv)/(dt)`

 `(dr)/(dt)=1/112.5xx1.5`

 `(dr)/(dt)=1/75`

Then the rate is `1/75` m/s

b. Let`(dr)/(dt)=1`

 `(dt)/(dr)=frac{1}{(dr)/(dt)}=1/0.1`

Use the chain rule 

 `(dv)/(dr)=(dv)/(dt)xx(dt)/(dr)`

 `(dv)/(dr)=1.5xx1/0.1`

 `(dv)/(dr)=15`

From (1)

 `15=9/2(r-1/2)^2`

 `(r-1/2)^2=10/3`

`r-1/2=sqrt(10/3)`or`r-1/2=-sqrt(10/3)`

`r=1/2+sqrt(10/3)`or `r=1/2-sqrt(10/3) <0`

Then `r=1/2+sqrt(10/3)`

Substitute `r=1/2+sqrt(10/3)` into `V=3/2(r-1/2)^3-1`

 `V=3/2(1/2+sqrt(10/3)-1/2)^3-1=8.13`

Then the volume is `8.13` m³ 

The equation of a curve is `y=3x+1-4(3x+1)^(1/2`  for `x> -1/3`

a. Find `(dy)/(dx)` and `(d^2y)/(dx^2)`

b. Find the coordinates of the stationary point of the curve and determine its nature.

a. Differentiate `y=3x+1-4(3x+1)^(1/2)`

 `(dy)/(dx)=3-4(1/2)(3)(3x+1)^(-1/2)=3-6(3x+1)^(-1/2)`

 `(d^2y)/(dx^2)=-6(-1/2)(3)(3x+1)^(-3/2)=9(3x+1)^(-3/2)`

b. For the curve has a stationary value

`3-6(3x+1)^(-1/2)=0`(as `(dy)/(dx)=0` )

 `3-frac{6}{sqrt(3x+1)}=0`

 `sqrt(3x+1)=2`

 `3x+1=4`

 `x=1`

Substitute `x=1` into `y=3x+1-4(3x+1)^(1/2)`

 `3(1)+1-4[3(1)+1]^(1/2)=-4`

The stationary point has coordinates `(1, −4)`

At `(1, −4)`

 `frac{d^2y}{dx^2}=9[3(1)+1]^(-3/2)=9/8`

`frac{d^2y}{dx^2}>0`then `(1, −4)` is a minimum point

Water is poured into a tank at a constant rate of `500` cm³ per second. The depth of water in the tank, `t` seconds after filling starts, is `h` cm. When the depth of water in the tank is`h` cm, the volume, `V` cm³, of water in the tank is given by the formula

`V=4/3(25+h)^3-62500/3` .

`a.`Find the rate at which `h` is increasing at the instant when `h=10` cm.

`b.` At another instant, the rate at which `h`  is increasing is `0.075` cm per second. Find the value of `V`  at this instant.

a. Differentiate `V=4/3(25+h)^3-62500/3`

`(dv)/(dh)=4/3(3)(25+h)^2=4(25+h)^2` (1) 

Let `h=10`

 `(dv)/(dh)=4(25+10)^2=4900`

 `(dh)/(dv)=1/{(dv)/(dh)}=1/4900`

Let `(dv)/(dt)=500`

Use the chain rule 

 `(dh)/(dt)=(dh)/(dv)xx(dv)/(dt)`

 `(dh)/(dt)=1/4900xx500`

 `(dh)/(dt)=0.102`

Then the rate is `0.102` cm/s

b. Let `(dh)/(dt)=0.075`

 `(dt)/(dh)=1/((dh)/(dt))=1/0.075`

Use the chain rule 

 `(dv)/(dh)=(dv)/(dt)xx(dt)/(dh)`

 `4(25+h)^2=500xx1/0.075`

 `4(25+h)^2=20000/12`

 `(25+h)^2=5000/3`

`25+h=sqrt(5000/3)` or `25+h=-sqrt(5000/3)`

`h=15.825` or `h=-65.825<0`

Then `h=15.825`

Substitute `h=15.825` into `V=4/3(25+h)^3-62500/3`

 `4/3(25+15.825)^3-62500/3=69889~~69900`

Then the volume is `69900` cm³