1) Differentiate the following with respect to x: a) In C 4x+3. 2x-5 [4] b) 3* Inx [5] c) Vx2 + 1 [4] d) e z [3] 2. Given that y = (a + bx)e- where a and b are constants, show that dx2 + 2- arty = 0 [8] 3. Find for 7x3 + 4y = 3y2 [4] dx 4. Given the equation 2y2 - 2xy - 4y + x2 = 0 of a curve, obtain the x-coordinates of each point at which the curve has a horizontal tangent. [6] 5. Find the gradient of cos(xy) - 3y4 + 3 at the point (1, ") [5] 6. Find the equation of the tangent to the curve xy + y3 = 6x + 3y at (1,2)

Providing step by step detailed solution for all parts of the question as shown below:

 

 

 

Step-by-step explanation

:::Question−1:::

For all parts from "a" to "d", we have to take the single derivative of respective functions,

 

Part-a

The function f(x) is defined as,

f(x)=ln(2x−54x+3?)

Performing differentiation operation of function f(x) with respect to x,

f′(x)=dxd?f(x)

f′(x)=dxd?ln(2x−54x+3?)

f′(x)=2x−54x+3?1?∗(2x−5)24(2x−5)−2(4x+3)?

f′(x)=4x+32x−5?∗(2x−5)28x−20−8x−6?

f′(x)=4x+32x−5?∗(2x−5)2−26?

f′(x)=−(4x+3)(2x−5)226(2x−5)?

 

 

Part-b

The function f(x) is defined as,

f(x)=3xln(x)

Performing differentiation operation of function f(x) with respect to x,

f′(x)=dxd?f(x)

f′(x)=dxd?3xln(x)

f′(x)=3xln(3)ln(x)+3x∗x1?

f′(x)=3xln(3)ln(x)+x3x?

 

 

Part-c

The function f(x) is defined as,

f(x)=x2+1?

f(x)=(x2+1)21?

Performing differentiation operation of function f(x) with respect to x,

f′(x)=dxd?f(x)

f′(x)=dxd?(x2+1)21?

f′(x)=21?(x2+1)21?−1∗(2x)

f′(x)=22x?(x2+1)−21?

f′(x)=22x?(x2+1)−21?

f′(x)=x(x2+1)−21?

f′(x)=(x2+1)21?x?

f′(x)=x2+1?x?

 

 

Part-d

The function f(x) is defined as,

f(x)=e−2x2?

Performing differentiation operation of function f(x) with respect to x,

f′(x)=dxd?f(x)

f′(x)=dxd?e−2x2?

f′(x)=e−2x2?∗(−22x?)

f′(x)=e−2x2?∗(−x)

f′(x)=−xe−2x2?

 

 

 

 

:::Question−2:::

The function y is described by,

y=(a+bx)e−x

Where, a and b are arbitrary constant.

Step-1) Take single derivative of function y with respect to x,

dxdy?=dxd?y

dxdy?=dxd?[(a+bx)e−x]

dxdy?=b∗e−x+(a+bx)∗(−1)∗e−x

dxdy?=be−x−(a+bx)e−x

dxdy?=be−x−ae−x−bxe−x

dxdy?=(b−a−bx)e−x

 

Step-2) Again we  will take the derivative of function dy/dx with respect to x,

dx2d2y?=dxd?(dxdy?)

dx2d2y?=dxd?((b−a−bx)e−x)

dx2d2y?=−be−x+(b−a−bx)e−x∗(−1)

dx2d2y?=−be−x−(b−a−bx)e−x

dx2d2y?=−be−x−be−x+ae−x+bxe−x

dx2d2y?=−2be−x+ae−x+bxe−x

dx2d2y?=(−2b+a+bx)e−x

 

Step-3) Now we will prove our equation,

LHS=dx2d2y?+2dxdy?+y

LHS=(−2b+a+bx)e−x+2∗(b−a−bx)e−x+(a+bx)e−x

LHS={(−2b+a+bx)+2∗(b−a−bx)+(a+bx)}e−x

LHS={−2b+a+bx+2b−2a−2bx+a+bx}e−x

LHS={−2b+2b−2a+a+a+bx+bx−2bx}e−x

LHS={−2b+2b−2a+2a+2bx−2bx}e−x

LHS={0+0+0}e−x

LHS=0∗e−x

LHS=0=RHS

Hence, we have proved that 

dx2d2y?+2dxdy?+y=0

 

 

 

 

:::Question−3:::

Given the function is expressed as,

7x3+4y=3y2

Taking derivative both the sides of equation with respect to x,

dxd?(7x3+4y)=dxd?(3y2)

dxd?(7x3)+dxd?(4y)=dxd?(3y2)

7dxd?(x3)+4dxd?(y)=3dxd?(y2)

7∗3x2+4dxdy?=3∗2y∗dxdy?

21x2+4dxdy?=6ydxdy?

21x2=6ydxdy?−4dxdy?

21x2=(6y−4)dxdy?

dxdy?=6y−421x2?

 

 

 

 

:::Question−4:::

Given the function is expressed as,

2y2−2xy−4y+x2=0

Taking derivative both the sides of equation with respect to x,

2∗2y∗dxdy?−2y−2xdxdy?−4∗dxdy?+2x=0

4ydxdy?−2y−2xdxdy?−4dxdy?+2x=0

(4y−2x−4)dxdy?−2y+2x=0

(4y−2x−4)dxdy?=2y−2x

dxdy?=4y−2x−42y−2x?

dxdy?=2y−x−2y−x?

 

When the tangent is horizontal, it means the derivative must must be equal to zero,

dxdy?=2y−x−2y−x?=0

2y−x−2y−x?=0

y−x=0

y=x

 

Substituting value of y=x in our equation,

2x2−2x∗x−4x+x2=0

2x2−2x2−4x+x2=0

−4x+x2=0

x(x−4)=0

x=0

x=4

 

Hence, at x = 0 and x = 4, the curve will have horizontal tangent

 

 

 

 

:::Question−5:::

Given the two variable function f(x,y) is expressed as,

f(x,y)=cos(xy)−3y4+3

Taking the gradient of function f(x,y)

?f(x,y)=∂x∂f(x,y)?ax??+∂y∂f(x,y)?ay??

?f(x,y)=−ysin(xy)ax??+(−xsin(xy)−12y3)ay??

 

Now we need to put x=1 and y=π/2 in above relation,

?f(x,y)=−2π?sin(1∗2π?)ax??+(−1sin(1∗2π?)−12∗(2π?)3)ay??

?f(x,y)=−2π?∗sin(2π?)ax??+(−1∗sin(2π?)−12∗(2π?)3)ay??

?f(x,y)=−2π?∗0 ax??+(−1∗0−12∗8π3?) ay??

?f(x,y)=0 ax??−23π3? ay??

 

 

 

 

:::Question−6:::

Given the function is expressed as,

xy2+y3=6x+3y

To get the equation of tangent, we need to take the derivative of above function with respect to x,

y2+2xydxdy?+3y2dxdy?=6+3dxdy?

(2xy−3+3y2)dxdy?=6−y2

dxdy?=2xy−3+3y26−y2?

Above is the equation of tangent. Here, we will put x=1 and y=2

dxdy?=2∗1∗2−3+3∗226−22?

dxdy?=4−3+3∗46−4?

dxdy?=4−3+126−4?

dxdy?=132?