Question

differentiate. g(x)=\frac{5x - 3}{8x + 5}+x^{3} g(x)=square

Explanation:

Step1: Apply sum - rule of differentiation

The sum - rule states that if \(y = u + v\), then \(y'=u'+v'\). Let \(u=\frac{5x - 3}{8x + 5}\) and \(v = x^{3}\). So \(g'(x)=u'+v'\).

Step2: Differentiate \(v = x^{3}\)

Using the power - rule \((x^{n})'=nx^{n - 1}\), for \(v=x^{3}\), we have \(v' = 3x^{2}\).

Step3: Differentiate \(u=\frac{5x - 3}{8x + 5}\) using the quotient - rule

The quotient - rule states that if \(u=\frac{f(x)}{g(x)}\), then \(u'=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^{2}}\). Here, \(f(x)=5x - 3\), \(f'(x)=5\), \(g(x)=8x + 5\), \(g'(x)=8\). Then \(u'=\frac{5(8x + 5)-(5x - 3)\times8}{(8x + 5)^{2}}=\frac{40x+25-(40x - 24)}{(8x + 5)^{2}}=\frac{40x + 25-40x + 24}{(8x + 5)^{2}}=\frac{49}{(8x + 5)^{2}}\).

Step4: Find \(g'(x)\)

Since \(g'(x)=u'+v'\), substituting \(u'\) and \(v'\) we get \(g'(x)=\frac{49}{(8x + 5)^{2}}+3x^{2}\).

Answer:

\(\frac{49}{(8x + 5)^{2}}+3x^{2}\)