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Find the derivative of each of these functions. 6. \( f(x)=-3 x^{12} \) 7. \( y=8 \pi^{3} \) 8. \( f(t)=t^{2}-t \) 9. \( y=\left(3 x^{2}+6\right)(2 x-1) \) 10. \( f(x)=e^{x}(2 x+\sin x) \) 11. \( y=\frac{4 x}{x^{2}+1} \) 12. \( y=\frac{x-2}{2 x^{2}-\cos x} \) 13. \( f(x)=\sin 5 x \) 14. \( f(x)=\log 2 x \) 15. \( f(x)=\left(x^{3}+2 x\right)^{27} \) 16. \( y=e^{x^{2}-1} \) 10 1 1.

Ask by Moreno Deleon. in Canada
Mar 14,2025

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1. \( f(x) = -3x^{12} \) Derivative: \( f'(x) = -36x^{11} \) 2. \( y = 8\pi^{3} \) Derivative: \( y' = 0 \) 3. \( f(t) = t^{2} - t \) Derivative: \( f'(t) = 2t - 1 \) 4. \( y = (3x^{2} + 6)(2x - 1) \) Derivative: \( y' = 18x^{2} - 6x + 12 \) 5. \( f(x) = e^{x}(2x + \sin x) \) Derivative: \( f'(x) = e^{x}(2x + 2 + \sin x + \cos x) \) 6. \( y = \frac{4x}{x^{2} + 1} \) Derivative: \( y' = \frac{4(1 - x^{2})}{(x^{2} + 1)^{2}} \) 7. \( y = \frac{x - 2}{2x^{2} - \cos x} \) Derivative: \( y' = \frac{(2x^{2} - \cos x) - (x - 2)(4x + \sin x)}{(2x^{2} - \cos x)^{2}} \) 8. \( f(x) = \sin 5x \) Derivative: \( f'(x) = 5\cos 5x \) 9. \( f(x) = \log (2x) \) Derivative: \( f'(x) = \frac{1}{x} \) 10. \( f(x) = (x^{3} + 2x)^{27} \) Derivative: \( f'(x) = 27(x^{3} + 2x)^{26}(3x^{2} + 2) \) 11. \( y = e^{x^{2} - 1} \) Derivative: \( y' = 2x e^{x^{2} - 1} \)

Solution

\( \textbf{6.}\quad f(x) = -3 x^{12} \) Derivative: \[ f'(x) = -3\cdot12\,x^{12-1} = -36\,x^{11} \] --- \( \textbf{7.}\quad y = 8 \pi^{3} \) Since \(8\pi^3\) is a constant, its derivative is \[ y' = 0 \] --- \( \textbf{8.}\quad f(t) = t^{2} - t \) Using the power rule: \[ f'(t) = 2t - 1 \] --- \( \textbf{9.}\quad y = \left(3 x^{2}+6\right)(2 x-1) \) Use the product rule where \[ u(x)=3x^2+6,\quad u'(x)=6x,\quad v(x)=2x-1,\quad v'(x)=2. \] Thus, \[ y' = u'(x)v(x)+u(x)v'(x)=6x(2x-1)+(3x^2+6)(2). \] Simplify: \[ 6x(2x-1)=12x^2-6x,\quad (3x^2+6)(2)=6x^2+12. \] So, \[ y' = (12x^2-6x)+(6x^2+12)=18x^2-6x+12. \] --- \( \textbf{10.}\quad f(x)=e^{x}(2 x+\sin x) \) Let \[ u(x)=e^x,\quad u'(x)=e^x,\quad v(x)=2x+\sin x,\quad v'(x)=2+\cos x. \] Apply the product rule: \[ f'(x)=u'(x)v(x)+u(x)v'(x)=e^x(2x+\sin x)+e^x(2+\cos x). \] Factor \(e^x\): \[ f'(x)=e^x\bigl((2x+\sin x)+(2+\cos x)\bigr)=e^x\bigl(2x+2+\sin x+\cos x\bigr). \] --- \( \textbf{11.}\quad y=\frac{4 x}{x^{2}+1} \) Use the quotient rule: \[ \text{If } y=\frac{u}{v},\quad y'=\frac{u'v-u\,v'}{v^2}. \] Here, \[ u(x)=4x,\quad u'(x)=4,\quad v(x)=x^2+1,\quad v'(x)=2x. \] Thus, \[ y'=\frac{4(x^2+1)-4x(2x)}{(x^2+1)^2}=\frac{4x^2+4-8x^2}{(x^2+1)^2}=\frac{4(1-x^2)}{(x^2+1)^2}. \] --- \( \textbf{12.}\quad y=\frac{x-2}{2 x^{2}-\cos x} \) Again, using the quotient rule with \[ u(x)=x-2,\quad u'(x)=1,\quad v(x)=2x^2-\cos x,\quad v'(x)=4x+\sin x. \] The derivative is: \[ y'=\frac{1\cdot (2x^2-\cos x)-(x-2)(4x+\sin x)}{(2x^2-\cos x)^2}. \] --- \( \textbf{13.}\quad f(x)=\sin 5x \) Using the chain rule: \[ f'(x)=\cos 5x\cdot 5=5\cos 5x. \] --- \( \textbf{14.}\quad f(x)=\log (2x) \) Assuming \(\log\) means the natural logarithm, use the chain rule: \[ f(x)=\ln (2x) \quad \longrightarrow \quad f'(x)=\frac{1}{2x}\cdot 2=\frac{1}{x}. \] --- \( \textbf{15.}\quad f(x)=\left(x^{3}+2 x\right)^{27} \) Let \( u(x)=x^3+2x \) so that \( f(x)=\bigl(u(x)\bigr)^{27} \). Then, \[ u'(x)=3x^2+2. \] By the chain rule, \[ f'(x)=27\bigl(u(x)\bigr)^{26}\cdot u'(x)=27\left(x^3+2x\right)^{26}(3x^2+2). \] --- \( \textbf{16.}\quad y=e^{x^{2}-1} \) Let \( u(x)=x^2-1 \) so that \( y=e^{u(x)} \) and \( u'(x)=2x \). Then, by the chain rule, \[ y' = e^{x^2-1}\cdot 2x=2x e^{x^2-1}. \]

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Bonus Knowledge

Derivatives can be fascinating pieces of mathematical art! For instance, in problem 6, applying the power rule gives \( f'(x) = -36 x^{11} \). Remember, the power rule states that you multiply by the exponent and decrease it by one! Another fun fact: in problem 10, you can use the product rule to find the derivative. Recall that for two functions multiplied together, you take the derivative of the first, multiply by the second, add the first multiplied by the derivative of the second. So, it becomes a beautiful dance of functions!

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