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Find the values of x and y for the given system of equations.
3x2y=3
2x+2y=4
The given system of equations can be expressed in the form of AX=B,
⇒ X= A^{1} B
We know that, A^{1}= (1/A) adj A
A^{1}= (1/10)
∴ X = A^{1} B=
Consider y=2(tanx)^{cotx}
Applying log in both sides,
logy=log2(tanx)^{cotx}
logy=log2+log(tanx)^{cotx}
logy=log2+cotx log(tanx)
Differentiating both sides with respect to x, we get
dy/dx = 2(tanx)^{cotx} (−csc^{2x}log(tanx)+cot^{2x}+1)
dy/dx = 2(tanx)^{cotx} (−csc^{2x}log(tanx)+csc^{2x})
dy/dx = 2(tanx)^{cotx} (csc^{2}x (1log(tanx))
∴ dy/dx =2 csc^{2}x.tanx^{cotx} (1log(tanx))
Given that, y=4x^{4}+2x
dy/dx = 16x^{3}+2
d^{2}y/dx^{2} =48x^{2}
48x^{2}−96x^{3}−12
=12(4x^{2}8x1)
Consider y=
y=5x^{3} (∴loge^{x}=x)
∴ dy/dx = 5(3x^{2})=15x^{2}
The total cost N(x) in rupees, associated with the production of x units of an item is given by N(x)=0.06x^{3}0.01x^{2}+10x43. Find the marginal cost when 5 units are produced.
The marginal cost is given by the rate of change of revenue.
Hence, (dN(x)/dt) =0.18x^{2}0.02x+10.
∴ = 0.18(5)^{2}0.02(5)+10
= 4.50.1+10
= Rs. 14.4
The cost of 8kg apple and 3kg is Rs 70. The cost of 10kg apple and 6kg orange is 90. Find the cost of each item if x is the cost of apples per kg and y is the cost of oranges per kg.
The given situation can be represented by a system of equations as:
8x+3y=70
10x+5y=90
Consider, A=
It can be expressed in the form of AX=B
⇒ X=A^{1} B
We know that, A^{1}=
∴ x=3, y=2.
i.e. The cost of apples is Rs 3 per kg and the cost of oranges is Rs 2 per kg.
Consider y=(3 cosx)^{x}
Applying log on both sides, we get
logy=log(3 cosx)^{x}
logy=x log(3 cosx)
logy=x(log3+log(cosx))
Differentiating both sides with respect to x, we get
log3+log(cosx)x tanx
dy/dx =y(log(3 cosx)x tanx)
dy/dx =(3 cosx)^{x} (log(3 cosx)x tanx)
Find the second order derivative y=e^{2x}+sin^{1}e^{x}.
Given that, y=e^{2x}+sin^{1}e^{x}
Consider y=7
The length of the rectangle is changing at a rate of 4 cm/s and the area is changing at the rate of 8 cm/s. What will be the rate of change of width if the length is 4cm and the width is 1 cm.
Let the length be l, width be b and the area be A.
The Area is given by A=lb
Given that, dl/dt =4cm/s and dA/dt =8 cm/s
Substituting in the above equation, we get
Given that, l=4 cm and b=1 cm
For a given system of equations if A=0 and (adj A)B≠O(zero matrix), then which of the following is correct regarding the solutions of the given equations?
If A is a singular matrix, then A=0
In this case, if (adj A) B≠O, then solution does not exist and the system of equations is called inconsistent.
Consider y=
Applying log to both sides, we get
logy=log
logy=
logy= (1/2) (log(x+1)log(3x1))
Differentiating with respect to x, we get
Find the second order derivative of y=3x^{2} 1 + log(4x)
Given that, y=3x^{2}+log(4x)
Consider logx
Differentiating w.r.t x by using chain rule, we get
For which of the values of x, the rate of increase of the function y=3x^{2}2x+7 is 4 times the rate of increase of x?
Given that,
y=3x^{2}2x+7
4=6x2
6x=6
⇒ x=1
Find the value of x and y for the given system of equations.
3x+4y=6
5x4y=4
The given system of equations can be expressed in the form of AX=B,
⇒X=A^{1} B
We know that, A^{1}=1/A adj A
Consider y=
Applying log on both sides, we get
logy=3e^{3x} logx
Differentiating both sides with respect to x, we get
Given that,
By using u.v rule, we get
Consider y=
Differentiating w.r.t x by using chain rule, we get
The volume of a cube of edge x is increasing at a rate of 12 cm/s. Find the rate of change of edge of the cube when the edge is 6 cm.
Let the volume of cube be V.
V=x^{3}
What is the slope of the tangent to the curve y = 2x/(x^{2} + 1) at (0, 0)?
We have y = 2x/(x^{2} + 1)
Differentiating y with respect to x, we get
dy/dx = d/dx(2x/(x^{2} + 1))
= 2 * [(x^{2} + 1)*1 – x * 2x]/(x^{2} + 1)^{2}
= 2 * [1 – x^{2}]/(x^{2} + 1)^{2}
Thus, the slope of tangent to the curve at (0, 0) is,
[dy/dx]_{(0, 0)} = 2 * [1 – 0]/(0 + 1)^{2}
Thus [dy/dx]_{(0, 0)} = 2.
Let y = √x. Let x=64 and Δx=0.3
Then, Δy=
dy is approximately equal to Δy is equal to:
dy=0.3/16=0.01875
∴ The approximate value of is 8+0.01875=8.01875
Using Integration by Substitution, Let xm=t
Differentiating w.r.t x, we get
mdx=dt
Replacing t with mx again we get,
∫8x^{3+1}dx
Using
∫8x^{3+1}dx=∫8x^{3}dx+∫1dx
=2x^{4}+x+C.
The given function can also be written as and is further used to solve integration by partial fractions numerical.
The value of f’(x) is 1 at the point P on a continuous curve y = f(x). What is the angle which the tangent to the curve at P makes with the positive direction of x axis?
Let, Φ be the angle which the tangent to the curve y = f(x) at P makes with the positive direction of the x axis.
Then,
tanΦ = [f’(x)]_{p} = 1
= tan(π/4)
So, it is clear that this can be written as,
= tan(π – π/4)
= tan(3π/4)
So, Φ = 3π/4
Therefore, the required angle which the tangent at P to the curve y = f(x) makes with positive direction of x axis is 3π/4.
Let y= Let x=49 and Δx=0.1
Then, Δy=
dy is approximately equal to Δy is equal to
dy=0.1/14=0.00714
∴ The approximate value of is 7+0.00714=7.00714
By using the method of integration by substitution,
Let x^{3}=t
Differentiating w.r.t x, we get
3x^{2} dx=dt
∫3x^{2}(cosx^{3}+8)dx=∫(cost+8)dt
∫(cost+8)dt=sint+8t
Replacing t with x^{3},we get
∫3x^{2}(cosx^{3}+8)dx=sinx^{3}+8x^{3}+C
To find ∫7x^{2}−x^{3}+2xdx
∫7x^{2}−x^{3}+2xdx=∫7x^{2}dx−∫x^{3}dx+2∫xdx
Using
As it is not proper rational function, we divide numerator by denominator and get
Let
So that, 5x–5 = A(x3) + B(x2)
Now, equating coefficients of x and constant on both sides, we get A + B = 5 and 3A + 2B = 5. Solving these equations, we get A=5 and B=10.
Therefore,
= x – 5logx2 + 10logx3+C
What will be the differential function of √(x^{2} + 2)?
Let, y = f(x) = √(x^{2} + 2)
So, f(x) = (x^{2} + 2)^{1/2}
On differentiating it we get,
f’(x) = d/dx[(x^{2} + 2)^{1/2}]
f’(x) = 1/2 * 1/√(x^{2} + 2) * 2x
So f’(x) = x/√(x^{2} + 2)
So the differential equation is:
dy = f’(x)dx
Hence, dy = x/√(x^{2} + 2) dx
Find the approximate value of f(5.03), where f(x)=4x^{2}7x+2.
Let x=5 and Δx=0.03
Then, f(x+Δx)=4(x+Δx)^{2}7(x+Δx)+2
Δy=f(x+Δx)f(x)
∴f(x+Δx)=Δy+f(x)
Δy=f’ (x)Δx
⇒ f(x+Δx)=f(x)+f’ (x)Δx
f(5.03)=(4(5)^{2}7(5)+2)+(8(5)7)(0.03) (∵ f’ (x)=8x7)
f(5.03)=(10035+2)+(407)(0.03)
f(5.03)=67+33(0.03)
f(5.03)=67+0.99=67.99
Let x^{2}=t
Differentiating w.r.t x, we get
2x dx=dt
∫6x(x^{2}+6)dx=3∫(t+6)dt
3∫(t+6)dt=3
Replacing t with x^{2}
∫6x(x^{2}+6)dx= +18x^{2}+C
To find ∫ 2 sin2x+3 dx
∫2sin2x+3dx=∫2sin2xdx+∫3dx
∫2sin2x+3dx=2∫sin2xdx+3∫dx
∫2sin2x+3dx= + 3x
∴∫2 sin2x+3 dx = cos2x+3x+C
It is a proper rational function. Therefore,
Where real numbers are determined, 1 = A(x+2) + B(x+1), Equating coefficients of x and the constant term, we get A+B = 0 and 2A+B = 1.
Solving it we get A=1, and B=1.
Thus, it simplifies to,
= logx+1 – logx+2 + C
What will be the differential function of log(x^{2} + 4)?
Let, y = f(x) = log(x^{2} + 4)
So f(x) = log(x^{2} + 4)
On differentiating it we get,
f’(x) = d/dx[log(x^{2} + 4)]
So f’(x) = 2x/(x^{2} + 4)
So the differential equation is:
dy = f’(x)dx
Hence, dy = 2x/(x^{2} + 4) dx
Let y=√x. Let x=9 and Δx=2
Then, Δy=
dy is approximately equal to Δy is equal to
dy=2/6=0.34
∴ The approximate value of √11 is 3+0.34=3.34.
Let logx=t
Differentiating w.r.t x, we get
Replacing t with logx, we get
To find
An improper integration fraction is reduced to proper fraction by _____
An improper integration factor can be reduced to proper fraction by division, i.e., if the numerator and denominator have same degree, then they must be divided in order to reduce it to proper fraction.
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