finding a derivative in exercises 7 - 26, use the rules of differentiation to find the derivative of the…

finding a derivative in exercises 7 - 26, use the rules of differentiation to find the derivative of the function.\n7. y = 12\n8. f(x)= - 9\n9. y = x^7\n10. y = x^12\n11. y = \\frac{1}{x^5}\n12. y = \\frac{3}{x^7}\n13. f(x)=\\sqrt9{x}\n14. g(x)=\\sqrt4{x}\n15. f(x)=x + 11\n16. g(x)=6x + 3\n17. f(t)= - 3t^2+2t - 4\n18. y = t^2-3t + 1\n19. g(x)=x^2 + 4x^3\n20. y = 4x-3x^3\n21. s(t)=t^3 + 5t^2-3t + 8\n22. y = 2x^3+6x^2 - 1\n23. y = \\frac{\\pi}{2}\\sin\\theta\n24. g(t)=\\pi\\cos t\n25. y = x^2-\\frac{1}{2}\\cos x\n26. y = 7x^4+2\\sin x

finding a derivative in exercises 7 - 26, use the rules of differentiation to find the derivative of the function.\n7. y = 12\n8. f(x)= - 9\n9. y = x^7\n10. y = x^12\n11. y = \\frac{1}{x^5}\n12. y = \\frac{3}{x^7}\n13. f(x)=\\sqrt9{x}\n14. g(x)=\\sqrt4{x}\n15. f(x)=x + 11\n16. g(x)=6x + 3\n17. f(t)= - 3t^2+2t - 4\n18. y = t^2-3t + 1\n19. g(x)=x^2 + 4x^3\n20. y = 4x-3x^3\n21. s(t)=t^3 + 5t^2-3t + 8\n22. y = 2x^3+6x^2 - 1\n23. y = \\frac{\\pi}{2}\\sin\\theta\n24. g(t)=\\pi\\cos t\n25. y = x^2-\\frac{1}{2}\\cos x\n26. y = 7x^4+2\\sin x

Answer

Explanation:

Step1: Recall power - rule and constant - rule

The power - rule for differentiation is $\frac{d}{dx}(x^n)=nx^{n - 1}$, and the derivative of a constant $C$ is $\frac{d}{dx}(C)=0$.

Step2: Differentiate $f(t)=-3t^{2}+2t - 4$

For the first term $-3t^{2}$: Using the power - rule, $\frac{d}{dt}(-3t^{2})=-3\times2t^{2 - 1}=-6t$. For the second term $2t$: Using the power - rule, $\frac{d}{dt}(2t)=2\times1t^{1 - 1}=2$. For the third term $-4$: Using the constant - rule, $\frac{d}{dt}(-4)=0$.

Step3: Combine the derivatives of each term

$f^\prime(t)=\frac{d}{dt}(-3t^{2}+2t - 4)=-6t + 2+0=-6t + 2$.

Answer:

$f^\prime(t)=-6t + 2$