how will you multiply
y = ae^3x + be^-2x with dy/dx = 3ae^3x - 2be^-2x
Strange,
you are taking Calculus and ask how to multiply
(ae^(3x) + be^(-2x))(3ae^(3x) - 2be^(-2x) ) ??
= 3a^2 e^(6x) - 2ab + 3ab - 2b^2 e^(-4x)
= 3a^2 e^(6x) + ab - 2b^2 e^(-4x)
Reiny...Just because Akash is in Calculus doensn't mean he/she cant ask how to multiply. U don't even know him/her so don't be judging!
To multiply the expression y = ae^3x + be^-2x with dy/dx = 3ae^3x - 2be^-2x, you can use the product rule of differentiation.
The product rule states that if you have two functions u(x) and v(x) and you want to differentiate their product, which is given by y(x) = u(x) * v(x), then the derivative of y with respect to x, dy/dx, can be found using the formula:
dy/dx = u'(x) * v(x) + u(x) * v'(x),
where u'(x) and v'(x) represent the derivatives of u(x) and v(x) with respect to x, respectively.
Let's apply the product rule to the given expression:
We have y = ae^3x + be^-2x.
First, we need to find the derivatives of ae^3x and be^-2x separately:
• For ae^3x, its derivative with respect to x is obtained by applying the chain rule since the base of the exponential function is changing with respect to x:
(d/dx) (ae^3x) = a * (d/dx) (e^3x) = a * 3e^3x = 3ae^3x.
• For be^-2x, we can directly apply the chain rule:
(d/dx) (be^-2x) = b * (d/dx) (e^-2x) = b * (-2e^-2x) = -2be^-2x.
Now, we can use the product rule to find dy/dx:
dy/dx = (d/dx) (ae^3x) * (be^-2x) + (ae^3x) * (d/dx) (be^-2x)
= 3ae^3x * be^-2x + ae^3x * (-2be^-2x)
= 3abe^(3x - 2x) + (-2ab)e^(3x - 2x)
= 3abe^x - 2abe^x
= (3a - 2a)be^x
= a(be^x).
Therefore, dy/dx = a(be^x).
Note: It is important to check the correctness of the calculation and the applicability of the product rule to ensure the accuracy of the result.