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Question

Question: If the derivative of \[\left( {ax - 5} \right){e^{3x}}\] at \[x = 0\] is \[ - 13\], then the value o...

If the derivative of (ax5)e3x\left( {ax - 5} \right){e^{3x}} at x=0x = 0 is 13 - 13, then the value of aais equal to

(A) 8 (B) - 5 (C) 5 (D) - 2 (E) 2  (A){\text{ 8}} \\\ (B){\text{ - 5}} \\\ (C){\text{ 5}} \\\ (D){\text{ - 2}} \\\ (E){\text{ 2}} \\\
Explanation

Solution

Hint:- Use the product rule to find derivatives.

Let, y=(ax5)e3x = \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=0x = 0is 13 - 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,
(d(uv)dx)=udvdx+vdudx\Rightarrow \left( {\dfrac{{d(uv)}}{{dx}}} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}
So, here u=e3x = {e^{3x}} and v=(ax5) = \left( {ax - 5} \right)
So, differentiating equation 1 with respect to x. We get,
dydx=e3xd(ax5)dx+(ax5)d(e3x)dx\Rightarrow \dfrac{{dy}}{{dx}} = {e^{3x}}\dfrac{{d(ax - 5)}}{{dx}} + (ax - 5)\dfrac{{d({e^{3x}})}}{{dx}} (By using product rule)
dydx=e3x(a)+(ax5)3e3x\Rightarrow \dfrac{{dy}}{{dx}} = {e^{3x}}(a) + (ax - 5)3{e^{3x}}
Now, putting x=0x = 0 in the above equation. We get,
(dydx)x=0=e0(a)+(a(0)5)3e0\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=0x = 0 is 13 - 13. So,
13=a15\Rightarrow - 13 = a - 15
a=2\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.