Question

let $f(x + y) = f(x)f(y)$ for all $x$ and $y$ , suppose $f(5) = 2$ and $f'(0) = 3$ then $f'(5)$ is
A.$6$
B.$7$
C.$4$
D.$8$

Answer 1

Hint : In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (output value) with respect to a change in its argument (input value). The derivative of a function of a single variable at a chosen input value, when it exists, is the slope of the tangent line to the graph of the function at that point. The tangent line is the best linear approximation of the function near that input value.

Complete step-by-step answer :
Assume that if $f$ is a real function and $x$ is a point in its domain. The derivative of $f$ at $x$ is defined by

\\\ f(x + h) - f(x) \\\ \end{gathered} }{h}$$ We have $$f(x + y) = f(x)f(y)$$ Putting $$x = 0$$ and $$y = 5$$ in the given equation , we get $$f(0 + 5) = f(0)f(5)$$ We have , $$f(5) = f(0)f(5)$$ Therefore we get $$f(0) = 1$$ Now consider $$f'(5) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(5 + h) - f(5)}}{h}$$ $$ = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(5)f(h) - f(5)}}{h}$$ Hence we get , $$f'(5) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(5)\left[ {f(h) - 1} \right]}}{h}$$ On simplifying we get , $$f'(5) = f(5)\mathop {\lim }\limits_{h \to 0} \dfrac{{\left[ {f(h) - 1} \right]}}{h}$$ Since we have found that $$f(0) = 1$$ from the above calculations We get , $$f'(5) = f(5)\mathop {\lim }\limits_{h \to 0} \dfrac{{\left[ {f(h) - f(0)} \right]}}{h}$$ Which simplifies to , $$f'(5) = f(5)f'(0)$$ Therefore we have , $$f'(5) = 2 \times 3 = 6$$ Therefore option (1) is the correct answer. **So, the correct answer is “Option 1”.** **Note:** The derivative of a function of a single variable at a chosen input value, when it exists, is the slope of the tangent line to the graph of the function at that point. Derivatives can be generalized to functions of several real variables . Differentiation is the action of computing a derivative of a given function .