ind d2ydx2 if y=2x4+5x3−3x−2 .
To find the second derivative of y = 2x^4 + 5x^3 - 3x - 2, we need to find the derivative of the first derivative.
First, we find the first derivative:
dy/dx = d/dx (2x^4 + 5x^3 - 3x - 2)
= 8x^3 + 15x^2 - 3
Now we find the second derivative by taking the derivative of the first derivative:
d^2y/dx^2 = d/dx (8x^3 + 15x^2 - 3)
= 24x^2 + 30x
Therefore, the second derivative of y = 2x^4 + 5x^3 - 3x - 2 is d^2y/dx^2 = 24x^2 + 30x.
To find the second derivative of the function y = 2x^4 + 5x^3 - 3x - 2, we need to differentiate it twice with respect to x.
First, let's find the first derivative:
y' = d/dx (2x^4 + 5x^3 - 3x - 2)
To find the derivative of each term, we can use the power rule of differentiation.
The power rule states that if we have a term of the form ax^n, where a is a constant, the derivative is given by:
d/dx (ax^n) = nax^(n-1)
Applying the power rule to each term, we get:
y' = (d/dx) (2x^4) + (d/dx) (5x^3) - (d/dx) (3x) - (d/dx) (2)
y' = 8x^3 + 15x^2 - 3
Now, let's find the second derivative by taking the derivative of y' with respect to x:
y'' = d/dx (8x^3 + 15x^2 - 3)
Again, applying the power rule:
y'' = (d/dx) (8x^3) + (d/dx) (15x^2) - (d/dx) (3)
y'' = 24x^2 + 30x
Therefore, the second derivative of y = 2x^4 + 5x^3 - 3x - 2 is y'' = 24x^2 + 30x
ind d2ydx2 if y=2x4+5x3−3x^−2 .
To find the second derivative of the function y=2x^4+5x^3−3x−2, follow these steps:
Step 1: Find the first derivative of the function.
To find the first derivative dy/dx, apply the power rule and sum rule of differentiation:
dy/dx = d(2x^4)/dx + d(5x^3)/dx + d(-3x)/dx + d(-2)/dx
dy/dx = 8x^3 + 15x^2 - 3
Step 2: Find the second derivative of the function.
To find the second derivative d^2y/dx^2, differentiate the first derivative with respect to x:
d^2y/dx^2 = d(8x^3 + 15x^2 - 3)/dx
d^2y/dx^2 = d(8x^3)/dx + d(15x^2)/dx + d(-3)/dx
d^2y/dx^2 = 24x^2 + 30x
Therefore, the second derivative of the function y=2x^4+5x^3−3x−2 is d^2y/dx^2 = 24x^2 + 30x.
To find the second derivative of y = 2x^4 + 5x^3 - 3x^(-2), we need to differentiate it twice with respect to x.
First, let's find the first derivative:
y' = d/dx (2x^4 + 5x^3 - 3x^(-2))
To find the derivative of each term, we can use the power rule of differentiation and the negative power rule.
For the term 2x^4, the power rule gives us:
(d/dx) (2x^4) = 8x^3
For the term 5x^3, the power rule gives us:
(d/dx) (5x^3) = 15x^2
For the term -3x^(-2), the negative power rule gives us:
(d/dx) (-3x^(-2)) = 6x^(-3)
So, the first derivative becomes:
y' = 8x^3 + 15x^2 + 6/x^3
Now, let's find the second derivative by taking the derivative of y' with respect to x:
y'' = d/dx (8x^3 + 15x^2 + 6/x^3)
For the term 8x^3, the power rule gives us:
(d/dx) (8x^3) = 24x^2
For the term 15x^2, the power rule gives us:
(d/dx) (15x^2) = 30x
For the term 6/x^3, we can use the negative power rule:
(d/dx) (6/x^3) = -18/x^4
So, the second derivative becomes:
y'' = 24x^2 + 30x - 18/x^4
Therefore, the second derivative of y = 2x^4 + 5x^3 - 3x^(-2) is y'' = 24x^2 + 30x - 18/x^4.
Find dydx if y=3x^−4(x+3) .
To find dy/dx for the function y = 3x^(-4)(x+3), we can use the product rule and the power rule of differentiation.
The product rule states that if we have two functions u(x) and v(x), the derivative of their product is given by:
(d/dx)(u(x)v(x)) = u'(x)v(x) + u(x)v'(x)
Applying the product rule to our function y = 3x^(-4)(x+3), we have:
u(x) = 3x^(-4)
v(x) = (x+3)
u'(x) = d/dx (3x^(-4))
= -12x^(-5)
v'(x) = d/dx (x+3)
= 1
Now, let's use the product rule to find dy/dx:
dy/dx = u'(x)v(x) + u(x)v'(x)
= (-12x^(-5))(x+3) + (3x^(-4))(1)
To simplify this expression, we can expand and combine like terms:
dy/dx = -12x^(-5)(x+3) + 3x^(-4)
= -12x^(-4) - 36x^(-5) + 3x^(-4)
Combining the terms with the same exponent of x, we get:
dy/dx = -9x^(-4) - 36x^(-5)
Therefore, the derivative of y = 3x^(-4)(x+3) is dy/dx = -9x^(-4) - 36x^(-5).
What is the displacement after 3 seconds of a particle moving along a line if s(t)=t3+2t2 , where s is distance in feet, and t is time in seconds?
To find the displacement after 3 seconds of a particle moving along a line, we need to find the difference between the initial and final positions.
The position function is given by s(t) = t^3 + 2t^2, where s is the distance in feet and t is the time in seconds.
To find the initial position, substitute t = 0 into the position function:
s(0) = 0^3 + 2(0)^2
s(0) = 0
The initial position is 0 feet.
Next, to find the final position, substitute t = 3 into the position function:
s(3) = (3)^3 + 2(3)^2
s(3) = 27 + 18
s(3) = 45
The final position is 45 feet.
Finally, we can find the displacement by subtracting the initial position from the final position:
Displacement = Final position - Initial position
Displacement = 45 - 0
Displacement = 45 feet
Therefore, the displacement after 3 seconds of a particle moving along a line is 45 feet.