Differentiate each function
a) y = -3x^2 + 5x - 4
b) f(x) = 6/x - 3/(x^2)
c) f(x) = (3x^2 - 4x)(x^3 + 1)
the answers i got for each was:
a) y'= -6x + 5
b) f'(x)= -6/x^2 + 6/x^3
c) f'(x)= 15x^4 + 6x - 16x^3 - 4
please double check and let me know if these are correct, thank you.
All correct, well done!
thank you! can u help me on this question please
determine the coordinates of two points on the plane with equation 5x + 4y - 3z = 6
Any point on the plane will satisfy the equation.
Since there are three variables x,y and z, you can fix any two variables and solve for the third.
Example:
Assuming arbitrarily y=6, z=6
=>
5x+4(6)-3(6)=6
5x+24-18=6
5x=0
x=0
So (0,6,6) is a point on the plane.
thank you
You're welcome!
To differentiate each function, we need to find their respective derivatives. Let's go through each function and calculate their derivatives:
a) y = -3x^2 + 5x - 4:
To find the derivative, we differentiate term by term. The derivative of a constant is zero, and for polynomial terms, we apply the power rule. Here's how we differentiate each term:
The derivative of -3x^2 is -6x (using the power rule, we bring down the exponent and multiply it by the coefficient).
The derivative of 5x is 5 (using the power rule, the derivative of x is 1, and we ignore the coefficient).
The derivative of -4 is 0.
Combining these derivatives, we have: y' = -6x + 5.
b) f(x) = 6/x - 3/(x^2):
To find the derivative of this function, we again differentiate term by term. Here's how we differentiate each term:
The derivative of 6/x is -6/x^2 (using the power rule, bringing down the exponent -1 and multiplying it by the coefficient).
The derivative of 3/(x^2) is -6/x^3 (using the power rule, bringing down the exponent -2 and multiplying it by the coefficient).
Combining these derivatives, we have: f'(x) = -6/x^2 + 6/x^3.
c) f(x) = (3x^2 - 4x)(x^3 + 1):
To differentiate this function, we use the product rule. The product rule states that the derivative of the product of two functions is equal to the derivative of the first function times the second function, plus the first function times the derivative of the second function. Applying the product rule, let's differentiate each term:
The derivative of 3x^2 is 6x (using the power rule as before).
The derivative of -4x is -4 (using the power rule).
The derivative of x^3 is 3x^2 (using the power rule).
The derivative of 1 is 0 (a constant).
Combining these derivatives and applying the product rule, we have:
f'(x) = (6x)(x^3 + 1) + (3x^2 - 4x)(3x^2)
Simplifying further, we get: f'(x) = 15x^4 + 6x - 16x^3 - 4.
Therefore, your answers are correct:
a) y' = -6x + 5
b) f'(x) = -6/x^2 + 6/x^3
c) f'(x) = 15x^4 + 6x - 16x^3 - 4