
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
506.4k+ views
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
\[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{\begin{gathered}
\\
f(x + h) - f(x) \\
\end{gathered} }{h}\]
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 .
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
\[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{\begin{gathered}
\\
f(x + h) - f(x) \\
\end{gathered} }{h}\]
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 .
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

