# 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}} \\

\]

Last updated date: 24th Mar 2023

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Total views: 310.8k

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Views today: 3.88k

Answer

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310.8k+ 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.

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