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Chapter 3: Problem 68

Use any method to evaluate the derivative of the following functions. $$f(z)=z^{2}\left(e^{3 z}+4\right)-\frac{2 z}{z^{2}+1}$$

Short Answer

Expert verified
The derivative of the function is: \(f'(z) = 2ze^{3z}+8z+3z^3e^{3z} + \frac{6z^2+2}{(z^2+1)^2}\).

Step by step solution

01

Differentiate the polynomial term

Apply the product rule of differentiation to the first term, \(z^2\left(e^{3 z}+4\right)\). The product rule states that if \(f(z) = u(z) \cdot v(z)\), then \(f'(z) = u'(z)v(z) + u(z)v'(z)\). In this case, \(u(z)=z^2\) and \(v(z)=e^{3z}+4\). First, find the derivatives of these functions: $$u'(z)=\frac{d}{dz}(z^2) = 2z$$ $$v'(z)=\frac{d}{dz}(e^{3z}+4) = 3e^{3z}$$ Now apply the product rule: $$\frac{d}{dz}\left(z^2\left(e^{3 z}+4\right)\right)= 2z\left(e^{3 z}+4\right) + z^2(3e^{3z})$$
02

Differentiate the rational term

Apply the quotient rule of differentiation to the second term, \(-\frac{2 z}{z^{2}+1}\). The quotient rule states that if \(f(z) = \frac{u(z)}{v(z)}\), then \(f'(z) = \frac{u'(z)v(z)-u(z)v'(z)}{v^2(z)}\). Here, \(u(z)=-2z\) and \(v(z)=z^2+1\). First, find the derivatives of these functions: $$u'(z) = \frac{d}{dz}(-2z) = -2$$ $$v'(z) = \frac{d}{dz}(z^2+1) = 2z$$ Now apply the quotient rule: $$\frac{d}{dz}\left(-\frac{2 z}{z^{2}+1}\right)=-\frac{(-2)(z^2+1)-(-2z)(2z)}{(z^2+1)^2}$$
03

Combine the results

Now that we have found the derivatives for both terms, we can combine them to find the derivative of the whole function: $$f'(z) = \left(2z\left(e^{3 z}+4\right) + z^2(3e^{3z})\right) + \left(-\frac{(-2)(z^2+1)-(-2z)(2z)}{(z^2+1)^2}\right)$$
04

Simplify the expression

Finally, simplify the expression to get the final derivative of the function: $$f'(z) = 2ze^{3z}+8z + 3z^3e^{3z}+\frac{2z^2+2+4z^2}{(z^2+1)^2} = 2ze^{3z}+8z+3z^3e^{3z} + \frac{6z^2+2}{(z^2+1)^2}$$

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