Math Doubts

Evaluate $\dfrac{dy}{dx}$ if $y \,=\, e^{x+3\log_{e}{(x)}}$

In this differentiation problem, an algebraic equation is given in terms of two variables $x$ and $y$. The value of $y$ is equal to the Napier constant $e$ raised to the power of the sum of the variable $x$ and three times the natural logarithm of $x$. It is given that the derivative of $y$ should be calculated with respect to $x$. The differentiation of the variable $y$ can be done in two mathematical approaches. So, let’s learn each method with step by step procedure.

Differentiation by the formulas

In this method, a derivative rule is used for the differentiation of the variable $y$ by differentiating the give algebraic equation $y \,=\, e^{\displaystyle \, x+3\log_{e}{(x)}}$.

Prepare the equation for the differentiation

Check the given algebraic equation to know whether the given algebraic equation is possible to simplify. The left hand side expression is a variable and it is not possible to simplify it. The right hand side expression in the equation is a mathematical expression and let’s try to simplify it before finding the differentiation

$y \,=\, e^{\displaystyle \, x+3\log_{e}{(x)}}$

In the right hand side expression of the equation, the base of the exponential function is $e$ and two expressions are connected by the summation at exponent position. It is possible to split the whole exponential expression into the product of two exponential functions as per the product rule of exponents.

$\implies$ $y$ $\,=\,$ $e^{\displaystyle \, x} \times e^{\displaystyle \, 3\log_{e}{(x)}}$

The multiplying factor $3$ can be shifted to the exponent position of the variable $x$ in the logarithmic function by the power rule of logarithms.

$\implies$ $y$ $\,=\,$ $e^{\displaystyle \, x} \times e^{\displaystyle \, \log_{e}{\big(x^3\big)}}$

According to the fundamental rule of logarithms, the mathematical constant $e$ raised to the power of natural logarithm of $x$ cubed is equal to $x$ cubed.

$\implies$ $y$ $\,=\,$ $e^{\displaystyle \, x} \times x^3$

Find the derivative by the Product rule

Let’s start differentiating the given algebraic equation with respect to $x$.

$\implies$ $\dfrac{d}{dx}{\,(y)}$ $\,=\,$ $\dfrac{d}{dx}{\,\Big(e^{\displaystyle \, x} \times x^3\Big)}$

$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{d}{dx}{\,\Big(e^{\displaystyle \, x} \times x^3\Big)}$

This equation expresses that the derivative of $y$ with respect to $x$ can be calculated by finding the differentiation of the product of the natural exponential function in $x$ and $x$ cubed. Due to the multiplication of the functions, the derivative can be evaluated by using the product rule of differentiation.

$\,\,=\,$ $e^{\displaystyle \, x} \times \dfrac{d}{dx}{\,\big(x^3\big)}$ $+$ $x^3 \times \dfrac{d}{dx}{\,\big(e^{\displaystyle \, x}\big)}$

The derivative of $x$ cubed is evaluated by the power rule of derivatives and the derivative of natural exponential function can be calculated from the derivative rule of natural exponential function.

$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $e^{\displaystyle \, x} \times 3x^{\, 3-1}$ $+$ $x^3 \times e^{\displaystyle \, x}$

Simplify the mathematical expression

The differentiation of the variable $y$ with respect to $x$ is evaluated successfully. It is time to simplify the mathematical expression in the right hand side of the equation.

$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $e^{\displaystyle \, x} \times 3x^{\, 2}$ $+$ $x^3 \times e^{\displaystyle \, x}$

In both terms of the right hand side expression of the equation, the natural exponential function in terms of $x$ is commonly appearing. Hence, it can be taken out commonly from the terms.

$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $e^{\displaystyle \, x} \times \big(3x^{\, 2}+x^3\big)$

In the second factor of the right hand side expression, the factor $x$ squared is common in both terms. Hence, it can also be separated by taking the factor common from them.

$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $e^{\displaystyle \, x} \times \big(3x^{\, 2}+x^{\,2+1}\big)$

$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $e^{\displaystyle \, x} \times \big(3x^{\, 2}+x^{\,2} \times x^{\,1}\big)$

$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $e^{\displaystyle \, x} \times \big(3x^{\, 2}+x^{\,2} \times x\big)$

$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $e^{\displaystyle \, x} \times x^{\, 2} \times \big(3+x\big)$

$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{dy}{dx}$ $\,=\,$ $e^{\displaystyle \, x} x^{\, 2} \big(3+x\big)$

Differentiation by the Limits

The given algebraic equation is $y \,=\, e^{\displaystyle \, x+3\log_{e}{(x)}}$ and it can simplified as follows.

$\implies$ $y \,=\, e^{\displaystyle \, x} \times e^{\displaystyle \, 3\log_{e}{(x)}}$

$\implies$ $y \,=\, e^{\displaystyle \, x} \times e^{\displaystyle \, \log_{e}{(x^3)}}$

$\implies$ $y \,=\, e^{\displaystyle \, x} \times x^3$

$\,\,\,\therefore\,\,\,\,\,\,$ $y \,=\, e^{\displaystyle \, x} x^3$

Differentiation by the First Principle

According to the fundamental definition of the derivative, the derivative of a function can be defined in limit form as follows. This first principle can be used to find the differentiation of the variable $y$.

$\dfrac{d}{dx}{\,f(x)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{f(x+h)-f(x)}{h}}$

Take $y \,=\, f(x)$, then

$(1).\,\,\,$ $f(x) \,=\, e^{\displaystyle \, x} x^3$

$(2).\,\,\,$ $f(x+h) \,=\, e^{\displaystyle \, (x+h)} (x+h)^3$

Now, substitute them in the first principle of the differentiation to start the procedure of the differentiation.

$\implies$ $\dfrac{d}{dx}{\,\Big(e^{\displaystyle \, x} x^3\Big)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle \, (x+h)} (x+h)^3-e^{\displaystyle \, x} x^3}{h}}$

Now, let’s try the direct substitution method to find the limit of the algebraic function in rational form.

$\implies$ $\dfrac{d}{dx}{\,\Big(e^{\displaystyle \, x} x^3\Big)}$ $\,=\,$ $\dfrac{e^{\displaystyle \, (x+0)} (x+0)^3-e^{\displaystyle \, x} x^3}{0}$

$\implies$ $\dfrac{d}{dx}{\,\Big(e^{\displaystyle \, x} x^3\Big)}$ $\,=\,$ $\dfrac{e^{\displaystyle \, x} x^3-e^{\displaystyle \, x} x^3}{0}$

$\implies$ $\dfrac{d}{dx}{\,\Big(e^{\displaystyle \, x} x^3\Big)}$ $\,=\,$ $\dfrac{\cancel{e^{\displaystyle \, x} x^3}-\cancel{e^{\displaystyle \, x} x^3}}{0}$

$\implies$ $\dfrac{d}{dx}{\,\Big(e^{\displaystyle \, x} x^3\Big)}$ $\,=\,$ $\dfrac{0}{0}$

It is evaluated that the derivative of the given function is indeterminate. It clears that the direct substitution method is not recommendable to find the differentiation of the simplified function. Therefore, we have to find the limit of the function in an alternative method.

Simplify the function in rational form

$\dfrac{d}{dx}{\,\Big(e^{\displaystyle \, x} x^3\Big)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle \, (x+h)} (x+h)^3-e^{\displaystyle \, x} x^3}{h}}$

Now, let’s try to simplify the right hand side expression of the equation for calculating the derivative of the simplified function by the limits.

$\,\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle \, (x+h)} \times (x+h)^3-e^{\displaystyle \, x} \times x^3}{h}}$

Observe the both terms in the numerator of the rational expression, there is a natural exponential function commonly in both terms of the function. So, split the natural exponential function in the first term of the numerator.

$\,\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle \, x} \times e^{\displaystyle \, h} \times (x+h)^3-e^{\displaystyle \, x} \times x^3}{h}}$

Now, take the common factor out from the terms in the numerator of the rational function.

$\,\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle \, x} \times \Big(e^{\displaystyle \, h} \times (x+h)^3-x^3\Big)}{h}}$

$\,\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg(e^{\displaystyle \, x} \times \dfrac{e^{\displaystyle \, h} \times (x+h)^3-x^3}{h}\Bigg)}$

In this case, the mathematical constant $e$ raised to the power $x$ is constant. Hence, it can be taken out from the limit operation by the constant multiple rule of the limits.

$\,\,=\,$ $\displaystyle e^{\displaystyle \, x} \times \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle \, h} \times (x+h)^3-x^3}{h}}$

In the first term of the numerator, two terms are connected by a plus sign in the second factor and the sum of terms has an exponent of three. It can be expanded by the cube of sum of two terms formula.

$\,\,=\,$ $\displaystyle e^{\displaystyle \, x} \times \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle \, h} \times \Big(x^3+h^3+3xh(x+h)\Big)-x^3}{h}}$

$\,\,=\,$ $\displaystyle e^{\displaystyle \, x} \times \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle \, h} \times \Big(x^3+h^3+3x^2h+3xh^2\Big)-x^3}{h}}$

The second term in the numerator is $x$ cubed. There is an $x$ cube term in the second factor of the first term. Hence, let us try to take it common from the terms for simplifying the expression further.

$\,\,=\,$ $\displaystyle e^{\displaystyle \, x} \times \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle \, h} \times \Big(x^3+(h^3+3x^2h+3xh^2)\Big)-x^3}{h}}$

For our convenience, the expression in the second factor of the first term in the numerator can be split as two terms. Now, multiply the terms in the expression by its factor as per distributive property.

$\,\,=\,$ $\displaystyle e^{\displaystyle \, x} \times \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle \, h} \times x^3+e^{\displaystyle \, h} \times (h^3+3x^2h+3xh^2)-x^3}{h}}$

$\,\,=\,$ $\displaystyle e^{\displaystyle \, x} \times \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle \, h} \times x^3-x^3+e^{\displaystyle \, h} \times (h^3+3x^2h+3xh^2)}{h}}$

$\,\,=\,$ $\displaystyle e^{\displaystyle \, x} \times \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle \, h} \times x^3-x^3 \times 1+e^{\displaystyle \, h} \times (h^3+3x^2h+3xh^2)}{h}}$

Now, the factor $x$ cubed is a common factor in both terms of the numerator. Hence, it can be taken out from them commonly.

$\,\,=\,$ $\displaystyle e^{\displaystyle \, x} \times \large \lim_{h \,\to\, 0}{\normalsize \dfrac{x^3 \times (e^{\displaystyle \, h}-1)+e^{\displaystyle \, h} \times (h^3+3x^2h+3xh^2)}{h}}$

The above rational expression can be split as sum of two terms.

$\,\,=\,$ $\displaystyle e^{\displaystyle \, x} \times \large \lim_{h \,\to\, 0}{\normalsize \bigg(\dfrac{x^3 \times (e^{\displaystyle \, h}-1)}{h}}$ $+$ $\dfrac{e^{\displaystyle \, h} \times (h^3+3x^2h+3xh^2)}{h}\bigg)$

According to the sum rule of limits, the limit of the sum of functions is equal to the sum of their limits.

$\,\,=\,$ $\displaystyle e^{\displaystyle \, x} \times \bigg(\large \lim_{h \,\to\, 0}{\normalsize \dfrac{x^3 \times (e^{\displaystyle \, h}-1)}{h}}$ $+$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle \, h} \times (h^3+3x^2h+3xh^2)}{h}\bigg)}$

In the second term of the second factor, $h$ is a common factor in each term of the second factor in the numerator. The expression in the denominator is also $h$ and it can cancel the common factor. Hence, take $h$ common from all terms in the numerator.

$\,\,=\,$ $\displaystyle e^{\displaystyle \, x} \times \bigg(\large \lim_{h \,\to\, 0}{\normalsize \dfrac{x^3 \times (e^{\displaystyle \, h}-1)}{h}}$ $+$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle \, h} \times h \times (h^2+3x^2+3xh)}{h}\bigg)}$

$\,\,=\,$ $\displaystyle e^{\displaystyle \, x} \times \bigg(\large \lim_{h \,\to\, 0}{\normalsize \bigg(x^3 \times \dfrac{(e^{\displaystyle \, h}-1)}{h}\bigg)}$ $+$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle \, h} \times \cancel{h} \times (h^2+3x^2+3xh)}{\cancel{h}}\bigg)}$

In the first term of the second factor, $x$ cubed is a constant and it can be released from the limit operation by the constant multiple rule of the limits.

$\therefore\,\,\,$ $\dfrac{d}{dx}{\,\Big(e^{\displaystyle \, x} x^3\Big)}$ $\,=\,$ $\displaystyle e^{\displaystyle \, x} \times \bigg(x^3 \times \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle \, h}-1}{h}}$ $+$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Big(e^{\displaystyle \, h} \times (h^2+3x^2+3xh)\Big)\bigg)}$

Evaluate the limit of each function

The simplification process is completed and it is the right time to evaluate the limit of each function.

$\,\,=\,$ $\displaystyle e^{\displaystyle \, x} \times \bigg(x^3 \times \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle \, h}-1}{h}}$ $+$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Big(e^{\displaystyle \, h} \times (h^2+3x^2+3xh)\Big)\bigg)}$

In the second factor, the limit of function is equal to one as per the natural exponential limit rule and the limit of the second factor can be evaluated by the direct substitution.

$\,\,=\,$ $\displaystyle e^{\displaystyle \, x} \times \bigg(x^3 \times 1$ $+$ $e^{\displaystyle \, 0} \times \Big(0^2+3x^2+3x(0)\Big)\bigg)$

$\,\,=\,$ $\displaystyle e^{\displaystyle \, x} \times \bigg(x^3$ $+$ $1 \times \Big(0+3x^2+0\Big)\bigg)$

$\,\,=\,$ $\displaystyle e^{\displaystyle \, x} \times \big(x^3+1 \times 3x^2\big)$

$\,\,=\,$ $\displaystyle e^{\displaystyle \, x} \times \big(x^3+3x^2\big)$

$\,\,=\,$ $\displaystyle e^{\displaystyle \, x} \times \big(x^{2+1}+3 \times x^2\big)$

$\,\,=\,$ $\displaystyle e^{\displaystyle \, x} \times \big(x^2 \times x^1+3 \times x^2\big)$

$\,\,=\,$ $\displaystyle e^{\displaystyle \, x} \times \big(x^2 \times x+3 \times x^2\big)$

$\,\,=\,$ $\displaystyle e^{\displaystyle \, x} \times x^2 \times (x+3)$

$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{dy}{dx}$ $\,=\,$ $\displaystyle e^{\displaystyle \, x} x^2 (x+3)$

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