
How do you find the derivative of $G\left( x \right) = \int\limits_1^x {{e^{{t^2}}}} $ ?
Answer
510.3k+ views
Hint: In the given question, we are required to differentiate the definite integral function having limits from one to x. So, we will use the Leibnitz Theorem to solve the given problem. So, we first assume the given definite integral as a variable $y$ and then differentiate both sides of the equation using Leibnitz rule $\dfrac{d}{{dx}}\left[ {\int\limits_c^y {f\left( t \right)} } \right] = f\left( x \right)\dfrac{{dy}}{{dx}}$, where c is any constant.
Complete step by step answer:
So, we are given the definite integral function,
$G\left( x \right) = \int\limits_1^x {{e^{{t^2}}}} $
Now, we have to differentiate the given function. We know the Leibnitz rule for differentiation of an integral. According to Leibnitz rule, we have,
$\dfrac{d}{{dx}}\left[ {\int\limits_c^y {f\left( t \right)} } \right] = f\left( x \right)\dfrac{{dy}}{{dx}}$
$ \Rightarrow \dfrac{d}{{dx}}\left[ {G\left( x \right)} \right] = \dfrac{d}{{dx}}\left[ {\int\limits_1^x {{e^{{t^2}}}} } \right]$
$ \Rightarrow \dfrac{d}{{dx}}\left[ {G\left( x \right)} \right] = \left( {{e^{{x^2}}}} \right) \times \dfrac{{d\left( {{x^2}} \right)}}{{dx}}$
Now, we know the power rule of differentiation. So, we have, $\dfrac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{\left( {n - 1} \right)}}$. So, the derivative of ${x^2}$ with respect to x is $\left( {2x} \right)$. Hence, we get,
$ \Rightarrow \dfrac{d}{{dx}}\left[ {G\left( x \right)} \right] = \left( {{e^{{x^2}}}} \right) \times \left( {2x} \right)$
Multiplying he terms and simplifying the expression, we get,
$ \therefore \dfrac{d}{{dx}}\left[ {G\left( x \right)} \right] = 2x{e^{{x^2}}}$
Therefore, the derivative of $G\left( x \right) = \int\limits_1^x {{e^{{t^2}}}} $ with respect to x is $2x{e^{{x^2}}}$.
Note: We must know the Leibnitz rule for differentiating the integral function given to us. According to the Leibnitz rule, the derivative of an integral of a function is the function itself. Basically, the two cancel each other out like addition and subtraction. Furthermore, we just take the variable in the top limit of the integral, x, and substituting it into the function being integrated and applying the chain rule of differentiation. We must know the derivatives of the basic functions in order to get through the given problem.
Complete step by step answer:
So, we are given the definite integral function,
$G\left( x \right) = \int\limits_1^x {{e^{{t^2}}}} $
Now, we have to differentiate the given function. We know the Leibnitz rule for differentiation of an integral. According to Leibnitz rule, we have,
$\dfrac{d}{{dx}}\left[ {\int\limits_c^y {f\left( t \right)} } \right] = f\left( x \right)\dfrac{{dy}}{{dx}}$
$ \Rightarrow \dfrac{d}{{dx}}\left[ {G\left( x \right)} \right] = \dfrac{d}{{dx}}\left[ {\int\limits_1^x {{e^{{t^2}}}} } \right]$
$ \Rightarrow \dfrac{d}{{dx}}\left[ {G\left( x \right)} \right] = \left( {{e^{{x^2}}}} \right) \times \dfrac{{d\left( {{x^2}} \right)}}{{dx}}$
Now, we know the power rule of differentiation. So, we have, $\dfrac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{\left( {n - 1} \right)}}$. So, the derivative of ${x^2}$ with respect to x is $\left( {2x} \right)$. Hence, we get,
$ \Rightarrow \dfrac{d}{{dx}}\left[ {G\left( x \right)} \right] = \left( {{e^{{x^2}}}} \right) \times \left( {2x} \right)$
Multiplying he terms and simplifying the expression, we get,
$ \therefore \dfrac{d}{{dx}}\left[ {G\left( x \right)} \right] = 2x{e^{{x^2}}}$
Therefore, the derivative of $G\left( x \right) = \int\limits_1^x {{e^{{t^2}}}} $ with respect to x is $2x{e^{{x^2}}}$.
Note: We must know the Leibnitz rule for differentiating the integral function given to us. According to the Leibnitz rule, the derivative of an integral of a function is the function itself. Basically, the two cancel each other out like addition and subtraction. Furthermore, we just take the variable in the top limit of the integral, x, and substituting it into the function being integrated and applying the chain rule of differentiation. We must know the derivatives of the basic functions in order to get through the given problem.
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