Y = (x + 2)(x – 1) (x + 3) find \({dy\over dx}\)
3x2 + 8x + 1
Derivative of a Product of Linear Functions
To find the derivative \(\frac{dy}{dx}\) of the given function \(Y = (x + 2)(x - 1)(x + 3)\), we will first expand the expression for \(Y\) into a standard polynomial form. Once it's in polynomial form, we can easily differentiate each term.
Expanding the Function \(Y\)
First, let's multiply the first two factors: \((x + 2)(x - 1)\).
- We can use the FOIL method (First, Outer, Inner, Last) for this binomial multiplication:
- First: \(x \times x = x^2\)
- Outer: \(x \times (-1) = -x\)
- Inner: \(2 \times x = 2x\)
- Last: \(2 \times (-1) = -2\)
Combining these terms, we get:
\[ (x + 2)(x - 1) = x^2 - x + 2x - 2 = x^2 + x - 2 \]
Now, substitute this result back into the expression for \(Y\):
\[ Y = (x^2 + x - 2)(x + 3) \]
Next, we multiply this quadratic expression by the remaining linear factor \((x + 3)\). We will distribute each term from the first parenthesis to each term in the second parenthesis:
- Multiply \(x^2\) by \((x + 3)\): \(x^2(x + 3) = x^3 + 3x^2\)
- Multiply \(x\) by \((x + 3)\): \(x(x + 3) = x^2 + 3x\)
- Multiply \(-2\) by \((x + 3)\): \(-2(x + 3) = -2x - 6\)
Now, combine all these results to get the full expanded form of \(Y\):
\[ Y = (x^3 + 3x^2) + (x^2 + 3x) + (-2x - 6) \]
Combine the like terms (terms with the same power of \(x\)):
\[ Y = x^3 + (3x^2 + x^2) + (3x - 2x) - 6 \]
\[ Y = x^3 + 4x^2 + x - 6 \]
Calculating the Derivative \(\frac{dy}{dx}\)
With \(Y\) now in a simplified polynomial form (\(Y = x^3 + 4x^2 + x - 6\)), we can differentiate each term with respect to \(x\). We will use the power rule for differentiation, which states that if \(f(x) = ax^n\), then \(\frac{df}{dx} = nax^{n-1}\). Also, the derivative of a constant term is \(0\).
- Derivative of \(x^3\): Applying the power rule with \(a=1\) and \(n=3\), we get \(3 \times 1 \times x^{3-1} = 3x^2\).
- Derivative of \(4x^2\): Applying the power rule with \(a=4\) and \(n=2\), we get \(2 \times 4 \times x^{2-1} = 8x\).
- Derivative of \(x\): This is equivalent to \(x^1\). Applying the power rule with \(a=1\) and \(n=1\), we get \(1 \times 1 \times x^{1-1} = 1x^0 = 1 \times 1 = 1\).
- Derivative of \(-6\): Since \(-6\) is a constant, its derivative is \(0\).
Now, we combine the derivatives of each term to find the overall derivative \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{d}{dx}(x^3) + \frac{d}{dx}(4x^2) + \frac{d}{dx}(x) - \frac{d}{dx}(6) \]
\[ \frac{dy}{dx} = 3x^2 + 8x + 1 - 0 \]
\[ \frac{dy}{dx} = 3x^2 + 8x + 1 \]
Final Derivative Result
The derivative of the function \(Y = (x + 2)(x - 1)(x + 3)\) is \(\frac{dy}{dx} = 3x^2 + 8x + 1\).
Important Questions from Differential Equations
The Green's function for the differential equation \(\rm\frac{d^2x}{dt^2}\) + x = f(t), satisfying the initial conditions x(0) = \(\rm\frac{dx}{dt}\) (0) = 0, is
G(t, τ) = \(\begin{cases}0 & \text { for } \quad 0<\rm t<\tau \\ \sin (\rm t−\tau) & \text { for } \quad \rm t>\tau\end{cases}\)
The solution of the differential equation when the source f(t) = θ(t) (the Heaviside step function) is
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