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.

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$$ \textbf{6.}\quad f(x) = -3 x^{12} $$  
Derivative:  
$$
f'(x) = -3\cdot12\,x^{12-1} = -36\,x^{11}
$$

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$$ \textbf{7.}\quad y = 8 \pi^{3} $$  
Since $$8\pi^3$$ is a constant, its derivative is  
$$
y' = 0
$$

---

$$ \textbf{8.}\quad f(t) = t^{2} - t $$  
Using the power rule:  
$$
f'(t) = 2t - 1
$$

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$$ \textbf{9.}\quad y = \left(3 x^{2}+6\right)(2 x-1) $$  
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,  
$$
y' = (12x^2-6x)+(6x^2+12)=18x^2-6x+12.
$$

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$$ \textbf{10.}\quad f(x)=e^{x}(2 x+\sin x) $$  
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 $$e^x$$:  
$$
f'(x)=e^x\bigl((2x+\sin x)+(2+\cos x)\bigr)=e^x\bigl(2x+2+\sin x+\cos x\bigr).
$$

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$$ \textbf{11.}\quad y=\frac{4 x}{x^{2}+1} $$  
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,  
$$
y'=\frac{4(x^2+1)-4x(2x)}{(x^2+1)^2}=\frac{4x^2+4-8x^2}{(x^2+1)^2}=\frac{4(1-x^2)}{(x^2+1)^2}.
$$

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$$ \textbf{12.}\quad y=\frac{x-2}{2 x^{2}-\cos x} $$  
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:  
$$
y'=\frac{1\cdot (2x^2-\cos x)-(x-2)(4x+\sin x)}{(2x^2-\cos x)^2}.
$$

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$$ \textbf{13.}\quad f(x)=\sin 5x $$  
Using the chain rule:  
$$
f'(x)=\cos 5x\cdot 5=5\cos 5x.
$$

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$$ \textbf{14.}\quad f(x)=\log (2x) $$  
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,  
$$
f'(x)=27\bigl(u(x)\bigr)^{26}\cdot u'(x)=27\left(x^3+2x\right)^{26}(3x^2+2).
$$

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$$ \textbf{16.}\quad y=e^{x^{2}-1} $$  
Let $$ u(x)=x^2-1 $$ so that $$ y=e^{u(x)} $$ and $$ u'(x)=2x $$. Then, by the chain rule,  
$$
y' = e^{x^2-1}\cdot 2x=2x e^{x^2-1}.
$$
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: $$ y' = 2x e^{x^{2} - 1} $$

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