3. suppose that ( f(4)=2 ), ( g(4)=5 ), ( f^{prime}(4)=6 ), and ( g^{prime}(4)=-3 ). find ( h^{prime}(4) )…

Explanation:

Step1: Differentiate ( h(x) = 3f(x)+8g(x) )

Using the sum and constant - multiple rules of differentiation ((u + v)^\prime=u^\prime + v^\prime) and ((cf(x))^\prime = cf^\prime(x)). (h^\prime(x)=3f^\prime(x)+8g^\prime(x)) Substitute (x = 4): (h^\prime(4)=3f^\prime(4)+8g^\prime(4)) (h^\prime(4)=3\times6 + 8\times(-3)) (h^\prime(4)=18-24=-6)

Step2: Differentiate ( h(x)=f(x)g(x) )

Using the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = f(x)) and (v = g(x)) (h^\prime(x)=f^\prime(x)g(x)+f(x)g^\prime(x)) Substitute (x = 4): (h^\prime(4)=f^\prime(4)g(4)+f(4)g^\prime(4)) (h^\prime(4)=6\times5+2\times(-3)) (h^\prime(4)=30 - 6=24)

Step3: Differentiate ( h(x)=\frac{f(x)}{g(x)} )

Using the quotient rule (\left(\frac{u}{v}\right)^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}), where (u = f(x)) and (v = g(x)) (h^\prime(x)=\frac{f^\prime(x)g(x)-f(x)g^\prime(x)}{g^{2}(x)}) Substitute (x = 4): (h^\prime(4)=\frac{f^\prime(4)g(4)-f(4)g^\prime(4)}{g^{2}(4)}) (h^\prime(4)=\frac{6\times5-2\times(-3)}{5^{2}}=\frac{30 + 6}{25}=\frac{36}{25}=1.44)

Step4: Differentiate ( h(x)=\frac{g(x)}{f(x)+g(x)} )

Using the quotient rule (\left(\frac{u}{v}\right)^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}), where (u = g(x)) and (v=f(x)+g(x)) First, (u^\prime=g^\prime(x)) and (v^\prime=f^\prime(x)+g^\prime(x)) (h^\prime(x)=\frac{g^\prime(x)(f(x)+g(x))-g(x)(f^\prime(x)+g^\prime(x))}{(f(x)+g(x))^{2}}) Expand the numerator: (h^\prime(x)=\frac{g^\prime(x)f(x)+g^\prime(x)g(x)-g(x)f^\prime(x)-g(x)g^\prime(x)}{(f(x)+g(x))^{2}}=\frac{g^\prime(x)f(x)-g(x)f^\prime(x)}{(f(x)+g(x))^{2}}) Substitute (x = 4): (h^\prime(4)=\frac{(-3)\times2-5\times6}{(2 + 5)^{2}}=\frac{-6-30}{49}=\frac{-36}{49}\approx - 0.735)

Answer:

a. (h^\prime(4)=-6) b. (h^\prime(4)=24) c. (h^\prime(4)=\frac{36}{25}) d. (h^\prime(4)=-\frac{36}{49})