
If the derivative of \[\left( {ax - 5} \right){e^{3x}}\] at \[x = 0\] is \[ - 13\], then the value of \[a\]is equal to
\[
(A){\text{ 8}} \\
(B){\text{ - 5}} \\
(C){\text{ 5}} \\
(D){\text{ - 2}} \\
(E){\text{ 2}} \\
\]
Answer
609.6k+ views
Hint:- Use the product rule to find derivatives.
Let, y\[ = \left( {ax - 5} \right){e^{3x}}\] (1)
As given in the question that the value of derivative of y with respect to x at \[x = 0\]is \[ - 13\].
As, y is a function of x. So, we can get the derivative of y easily by using product rule.
Which states that if u and v are two functions then,
\[ \Rightarrow \left( {\dfrac{{d(uv)}}{{dx}}} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\]
So, here u\[ = {e^{3x}}\] and v\[ = \left( {ax - 5} \right)\]
So, differentiating equation 1 with respect to x. We get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{3x}}\dfrac{{d(ax - 5)}}{{dx}} + (ax - 5)\dfrac{{d({e^{3x}})}}{{dx}}\] (By using product rule)
\[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{3x}}(a) + (ax - 5)3{e^{3x}}\]
Now, putting \[x = 0\] in the above equation. We get,
\[ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{x = 0}} = {e^0}(a) + (a(0) - 5)3{e^0}\]
As, given in the question, the derivative of the given function i.e. y at \[x = 0\] is \[ - 13\]. So,
\[ \Rightarrow - 13 = a - 15\]
\[ \Rightarrow a = 2\]
Hence, the correct option is E.
Note:- Whenever we come up with this type of problem where we are given with a function and the value of the derivative of that function at a given point, we first calculate the derivative of that function at a known point, then equate it with the given value of the derivative of the function at that point to get the required value of the variable.
Let, y\[ = \left( {ax - 5} \right){e^{3x}}\] (1)
As given in the question that the value of derivative of y with respect to x at \[x = 0\]is \[ - 13\].
As, y is a function of x. So, we can get the derivative of y easily by using product rule.
Which states that if u and v are two functions then,
\[ \Rightarrow \left( {\dfrac{{d(uv)}}{{dx}}} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\]
So, here u\[ = {e^{3x}}\] and v\[ = \left( {ax - 5} \right)\]
So, differentiating equation 1 with respect to x. We get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{3x}}\dfrac{{d(ax - 5)}}{{dx}} + (ax - 5)\dfrac{{d({e^{3x}})}}{{dx}}\] (By using product rule)
\[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{3x}}(a) + (ax - 5)3{e^{3x}}\]
Now, putting \[x = 0\] in the above equation. We get,
\[ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{x = 0}} = {e^0}(a) + (a(0) - 5)3{e^0}\]
As, given in the question, the derivative of the given function i.e. y at \[x = 0\] is \[ - 13\]. So,
\[ \Rightarrow - 13 = a - 15\]
\[ \Rightarrow a = 2\]
Hence, the correct option is E.
Note:- Whenever we come up with this type of problem where we are given with a function and the value of the derivative of that function at a given point, we first calculate the derivative of that function at a known point, then equate it with the given value of the derivative of the function at that point to get the required value of the variable.
Recently Updated Pages
A man running at a speed 5 ms is viewed in the side class 12 physics CBSE

The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

State and explain Hardy Weinbergs Principle class 12 biology CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Which of the following statements is wrong a Amnion class 12 biology CBSE

Differentiate between action potential and resting class 12 biology CBSE

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

How much time does it take to bleed after eating p class 12 biology CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

Explain sex determination in humans with the help of class 12 biology CBSE

