Question

Find $ \dfrac{d}{{dx}}[{e^{( - a{x^2})}}\log \sin x = $
\[\left( 1 \right)\]${e^{( - a{x^2})}}[\cot x + 2ax \times \log \sin x]$
\[\left( 2 \right)\]${e^{( - a{x^2})}}[\cot x + ax \times \log \sin x]$
\[\left( 3 \right)\]${e^{( - a{x^2})}}[\cot x - 2ax \times \log \sin x]$
\[\left( 4 \right)\]\[none{\text{ }}of{\text{ }}these\]

Answer 1

Hint: We have to find the derivative of $[{e^{( - a{x^2})}}\log \sin x]$ with respect to. We solve this using chain rule and product rule of differentiation . Also using various basic derivative formulas of trigonometric functions , derivatives of exponential functions and derivatives of logarithmic functions . We firstly derivate the function with respect to x by applying the product rule and then the chain rule.

Complete step-by-step answer:
Derivative of sum of two function is equal to sum of the derivatives of the functions :
\[{\text{ }}\dfrac{{d\left[ {f\left( x \right){\text{ }} + {\text{ }}g\left( x \right){\text{ }}} \right]}}{{dx}} = \] \[\dfrac{{d{\text{ }}f\left( x \right)}}{{dx}}{\text{ }} + \dfrac{{d{\text{ }}g\left( x \right)}}{{dx}}\]
Derivative of product of two function is difference of the derivatives of the functions :
\[\dfrac{{d\left[ {f\left( x \right){\text{ }} - {\text{ }}g\left( x \right)} \right]}}{{dx}}{\text{ }} = \] \[\dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}}{\text{ }} - {\text{ }}\dfrac{{d\left[ {g\left( x \right)} \right]}}{{dx}}\]
Derivative of product of two function is given by the following product rule :
\[\dfrac{{d\left[ {f\left( x \right){\text{ }} \times {\text{ }}g\left( x \right)} \right]}}{{dx}}\] \[ = {\text{ }}\dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}}{\text{ }} \times {\text{ }}g{\text{ }} + {\text{ }}f{\text{ }} \times {\text{ }}\dfrac{{d\left[ {g\left( x \right)} \right]}}{{dx}}\]
Derivative of quotient of two function is given by the following quotient rule :
\[\dfrac{{d\left[ {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right]}}{{dx}}{\text{ }} = \] $\dfrac{{[\dfrac{{d[f(x)]}}{{dx}} \times g(x) - f(x) \times \dfrac{{d[g(x)]}}{{dx}}]}}{{{{[g(x)]}^2}}}$

Given : $\dfrac{d}{{dx}}[{e^{( - a{x^2})}}logsinx]$
Let us consider $y = [{e^{( - a{x^2})}}logsinx]$
Now we have to derivative \[y\]with respect to
Using the formula of product rule
\[\dfrac{{d\left[ {f\left( x \right){\text{ }} \times {\text{ }}g\left( x \right)} \right]}}{{dx}}{\text{ }} = \] \[{\text{ }}\dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx{\text{ }}}} \times {\text{ }}g{\text{ }} + {\text{ }}f{\text{ }} \times {\text{ }}\dfrac{{d\left[ {g\left( x \right)} \right]}}{{dx}}\]
Differentiate \[y\]with respect to, we get
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}[{e^{( - a{x^2})}}] \times \log \sin x + \dfrac{d}{{dx}}[\log \sin x] \times [{e^{( - a{x^2})}}]$
Using chain rule and derivatives of functions
We know ,
( Derivative of ${e^x} = {e^x}$)
( derivative of ${x^n} = n \times {x^{(n - 1)}}$)
( derivative of \[log{\text{ }}x{\text{ }} = {\text{ }}\dfrac{1}{x}\])
( Derivative of\[sin{\text{ }}x{\text{ }} = {\text{ }}cos{\text{ }}x\])
Using these derivatives , we get
\[{\text{ }}\dfrac{{dy}}{{dx}} = \] $[{e^{( - a{x^2})}} \times ( - 2ax)] \times \log \sin x + [\dfrac{{\cos x}}{{\sin x}}] \times [{e^{( - a{x^2})}}]$
Also , we know \[cot{\text{ }}x{\text{ }} = {\text{ }}\dfrac{{cos{\text{ }}x}}{{sin{\text{ }}x}}\]
So,
\[\dfrac{{dy}}{{dx}}{\text{ }} = \] $[{e^{( - a{x^2})}}( - 2ax)] \times \log \sin x + [\cot x][{e^{( - a{x^2})}}]$
Taking ${e^{( - a{x^2})}}$common , we get
\[\dfrac{{dy}}{{dx}}{\text{ }} = \] $[{e^{( - a{x^2})}}] \times [\cot x - 2ax \times \log \sin x]$
Thus , the correct option is \[\left( 3 \right)\]
So, the correct answer is “Option 3”.

Note: We differentiated $y$ with respect to to find\[\dfrac{{dy}}{{dx}}\]. We know the differentiation of trigonometric function :
$\dfrac{{d\left[ {cos{\text{ }}x} \right]}}{{dx}} = {\text{ }} - sin{\text{ }}x$
\[\dfrac{{d\left[ {sin{\text{ }}x} \right]}}{{dx}}{\text{ }} = {\text{ }}cos{\text{ }}x\]
$d[{x^n}] = n{x^{(n - 1)}}$
$d[\tan x] = se{c^2}x$
We use the derivative according to the given problem .