let $f(x)=sqrt{3x^{2}+4x + 7}$\n$f(x)=$\n$\frac{6x + 4}{2sqrt{3x^{2}+4x + 7}}$\n$f(4)=$\n$40.852$
let $f(x)=sqrt{3x^{2}+4x + 7}$\n$f(x)=$\n$\frac{6x + 4}{2sqrt{3x^{2}+4x + 7}}$\n$f(4)=$\n$40.852$
Answer
Explanation:
Step1: Identify the outer - inner functions
Let (u = 3x^{2}+4x + 7), then (y=\sqrt{u}=u^{\frac{1}{2}}).
Step2: Differentiate the outer function
Using the power rule ((x^{n})'=nx^{n - 1}), for (y = u^{\frac{1}{2}}), (y'_u=\frac{1}{2}u^{-\frac{1}{2}}).
Step3: Differentiate the inner function
For (u = 3x^{2}+4x + 7), (u'_x=6x + 4).
Step4: Apply the chain - rule
The chain - rule states that (f'(x)=y'_u\cdot u'_x). Substituting (y'_u) and (u'_x) we get (f'(x)=\frac{6x + 4}{2\sqrt{3x^{2}+4x + 7}}).
Step5: Evaluate (f'(4))
Substitute (x = 4) into (f'(x)): First, calculate the denominator: (\sqrt{3x^{2}+4x + 7}=\sqrt{3\times4^{2}+4\times4 + 7}=\sqrt{48+16 + 7}=\sqrt{71}). The numerator is (6x + 4=6\times4+4=24 + 4=28). Then (f'(4)=\frac{28}{2\sqrt{71}}=\frac{14}{\sqrt{71}}\approx\frac{14}{8.4261}\approx1.6615).
Answer:
(1.6615)