Question

let $f(x)=sqrt{3x^{2}+4x + 7}$
$f(x)=$
$\frac{6x + 4}{2sqrt{3x^{2}+4x + 7}}$
$f(4)=$
$40.852$

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:

\(1.6615\)