differentiate.\nf(x) = \\frac{x}{x + \\frac{c}{x}}\nf(x) = \n\nresources\nread it watch it\n\nsubmit…

Explanation:

Step1: Simplify the first - function

First, simplify $f(x)=\frac{x}{x + \frac{c}{x}}=\frac{x}{\frac{x^{2}+c}{x}}=\frac{x^{2}}{x^{2}+c}$.

Step2: Use the quotient - rule for differentiation

The quotient rule is $\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}$. Here, $u = x^{2}$, $u'=2x$, $v=x^{2}+c$, $v' = 2x$. Then $f'(x)=\frac{2x(x^{2}+c)-x^{2}(2x)}{(x^{2}+c)^{2}}=\frac{2x^{3}+2cx - 2x^{3}}{(x^{2}+c)^{2}}=\frac{2cx}{(x^{2}+c)^{2}}$.

Step3: Differentiate the second - function using the product rule

The product rule is $(uv)'=u'v + uv'$. For $f(x)=(x^{3}+4)e^{x}$, where $u = x^{3}+4$, $u'=3x^{2}$, $v = e^{x}$, $v'=e^{x}$. Then $f'(x)=3x^{2}e^{x}+(x^{3}+4)e^{x}=(x^{3}+3x^{2}+4)e^{x}$.

Step4: Differentiate $f'(x)$ for the second - derivative of the second - function

Again, using the product rule on $f'(x)=(x^{3}+3x^{2}+4)e^{x}$, with $u=x^{3}+3x^{2}+4$, $u'=3x^{2}+6x$, $v = e^{x}$, $v'=e^{x}$. Then $f''(x)=(3x^{2}+6x)e^{x}+(x^{3}+3x^{2}+4)e^{x}=(x^{3}+6x^{2}+6x + 4)e^{x}$.

Answer:

For $f(x)=\frac{x}{x+\frac{c}{x}}$, $f'(x)=\frac{2cx}{(x^{2}+c)^{2}}$; for $f(x)=(x^{3}+4)e^{x}$, $f'(x)=(x^{3}+3x^{2}+4)e^{x}$, $f''(x)=(x^{3}+6x^{2}+6x + 4)e^{x}$