Question

(a) The function k is defined by k(x)=f(x)g(x). Find k′(0).

Answer 1

Answer:

(a) k'(0) = f'(0)g(0) + f(0)g'(0)

(b) m'(5) = $\frac{f'(5)g(5) - f(5)g'(5)}{2g^{2}(5) }$

Step-by-step explanation:

(a) Since k(x) is a function of two functions f(x) and g(x) [ k(x)=f(x)g(x) ], so for differentiating k(x) we need to use product rule,i.e., $\frac{\mathrm{d} [f(x)\times g(x)]}{\mathrm{d} x}=\frac{\mathrm{d} f(x)}{\mathrm{d} x}\times g(x) + f(x)\times\frac{\mathrm{d} g(x)}{\mathrm{d} x}$

this will give k'(x)=f'(x)g(x) + f(x)g'(x)

on substituting the value x=0, we will get the value of k'(0)

{for expressing the value in terms of numbers first we need to know the value of f(0), g(0), f'(0) and g'(0) in terms of numbers}{If f(0)=0 and g(0)=0, and f'(0) and g'(0) exists then k'(0)=0}

(b) m(x) is a function of two functions f(x) and g(x) [ $m(x)=\frac{1}{2}\times\frac{f(x)}{g(x)}$ ]. Since m(x) has a function g(x) in the denominator so we need to use division rule to differentiate m(x). Division rule is as follows : $\frac{\mathrm{d} \frac{f(x)}{g(x)}}{\mathrm{d} x}=\frac{\frac{\mathrm{d} f(x)}{\mathrm{d} x}\times g(x) + f(x)\times\frac{\mathrm{d} g(x)}{\mathrm{d} x}}{g^{2}(x)}$

this will give $m'(x) = \frac{1}{2}\times\frac{f'(x)g(x) - f(x)g'(x)}{g^{2}(x) }$

on substituting the value x=5, we will get the value of m'(5).

{for expressing the value in terms of numbers first we need to know the value of f(5), g(5), f'(5) and g'(5) in terms of numbers}

{NOTE : in m(x), g(x) ≠ 0 for all x in domain to make m(x) defined and even m'(x) }

{ NOTE : $\frac{\mathrm{d} f(x)}{\mathrm{d} x}=f'(x)$ }