Using the Quotient Rule In Exercises 7-12, use the Quotient Rule to find the derivative of the function. \( \begin{array}{ll}\text { 7. } f(x)=\frac{x}{x^{2}+1} & \text { 8. } g(t)=\frac{3 t^{2}-1}{2 t+5} \\ \text { 9. } h(x)=\frac{\sqrt{x}}{x^{2}+1} & \text { 10. } f(x)=\frac{x^{2}}{2 \sqrt{x}+1} \\ \text { 11. } g(x)=\frac{\sin x}{x^{2}} & \text { 12. } f(t)=\frac{\cos t}{t^{3}}\end{array} \)

We are given functions written as quotients. Recall that the Quotient Rule states that if   F(x) = U(x)/V(x), then   F ′(x) = [V(x)·U ′(x) − U(x)·V ′(x)]⁄[V(x)]². Below are the step‐by‐step calculations for Exercises 7–12. ───────────────────────────── 1. Exercise 7:   f(x) = x⁄(x² + 1) Here, U(x) = x with U ′(x) = 1, and V(x) = x² + 1 with V ′(x) = 2x. Applying the Quotient Rule:   f ′(x) = [(x² + 1)(1) − x(2x)]⁄(x² + 1)²        = [x² + 1 − 2x²]⁄(x² + 1)²        = (1 − x²)⁄(x² + 1)² ───────────────────────────── 2. Exercise 8:   g(t) = (3t² − 1)⁄(2t + 5) Here, U(t) = 3t² − 1 with U ′(t) = 6t, and V(t) = 2t + 5 with V ′(t) = 2. Then,   g ′(t) = [(2t + 5)(6t) − (3t² − 1)(2)]⁄(2t + 5)²        = [12t² + 30t − 6t² + 2]⁄(2t + 5)²        = (6t² + 30t + 2)⁄(2t + 5)² ───────────────────────────── 3. Exercise 9:   h(x) = √x⁄(x² + 1)     (write √x as x^(1/2)) Let U(x) = x^(1/2) so that U ′(x) = (1)/(2x^(1/2)) and V(x) = x² + 1 so that V ′(x) = 2x. Thus,   h ′(x) = [(x² + 1)·(1/(2√x)) − √x·(2x)]⁄(x² + 1)² Now, simplify the numerator:   = [ (x² + 1)/(2√x) − 2x√x ]⁄(x² + 1)² Recognize that 2x√x = 2x^(3/2). To combine terms, find a common denominator (2√x):   = [ (x² + 1 − 4x²)/(2√x) ]⁄(x² + 1)²   = (1 − 3x²)⁄[2√x·(x² + 1)²] ───────────────────────────── 4. Exercise 10:   f(x) = x²⁄(2√x + 1)     (write √x as x^(1/2)) Let U(x) = x² so that U ′(x) = 2x, and let V(x) = 2x^(1/2) + 1. Differentiate V(x): since the derivative of 2x^(1/2) is 2·(1/2)x^(–1/2) = x^(–1/2), we have V ′(x) = 1/√x. Now apply the Quotient Rule:   f ′(x) = [(2√x + 1)(2x) − x²·(1/√x)]⁄(2√x + 1)² Simplify the numerator:   (2√x + 1)(2x) = 4x√x + 2x  and  x²·(1/√x) = x^(3/2) Thus,   Numerator = 4x^(3/2) + 2x − x^(3/2) = 3x^(3/2) + 2x We can factor an x:   = x(3√x + 2) So,   f ′(x) = [x(3√x + 2)]⁄(2√x + 1)² ───────────────────────────── 5. Exercise 11:   g(x) = sin x⁄x² Let U(x) = sin x so that U ′(x) = cos x, and V(x) = x² so that V ′(x) = 2x. Then,   g ′(x) = [x²·cos x − sin x·(2x)]⁄(x²)²        = [x² cos x − 2x sin x]⁄x⁴ Factor an x from the numerator:   = x(x cos x − 2 sin x)⁄x⁴ = (x cos x − 2 sin x)⁄x³ ───────────────────────────── 6. Exercise 12:   f(t) = cos t⁄t³ Here, U(t) = cos t with U ′(t) = −sin t, and V(t) = t³ with V ′(t) = 3t². Using the Quotient Rule:   f ′(t) = [t³·(−sin t) − cos t·(3t²)]⁄(t³)²        = [−t³ sin t − 3t² cos t]⁄t⁶ Factor t² from the numerator:   = −t² (t sin t + 3 cos t)⁄t⁶ = −(t sin t + 3 cos t)⁄t⁴ ───────────────────────────── Final Answers: 7. f ′(x) = (1 − x²)⁄(x² + 1)² 8. g ′(t) = (6t² + 30t + 2)⁄(2t + 5)² 9. h ′(x) = (1 − 3x²)⁄[2√x · (x² + 1)²] 10. f ′(x) = [x(3√x + 2)]⁄(2√x + 1)² 11. g ′(x) = (x cos x − 2 sin x)⁄x³ 12. f ′(t) = −(t sin t + 3 cos t)⁄t⁴ Any answer that is algebraically equivalent to these is correct.