# Find the derivative of the following: - \[y={{2}^{x}}-3{{e}^{x}}-{{4}^{x}}\]

Answer

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Hint: Use basic formulae for derivative of \[{{a}^{x}},{{x}^{a}}\And {{e}^{x}}\].

We have expression/function

\[y={{2}^{x}}-3{{e}^{x}}-{{4}^{x}}-(1)\]

Now, let us differentiate the given function as,

\[\dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{2}^{x}}-3{{e}^{x}}-{{4}^{x}} \right)\]

One important rule should be applied here as stated below:

If we have $n$ functions ${{f}_{1}}(x),{{f}_{2}}(x),{{f}_{3}}(x)......{{f}_{n}}(x)$ and ${{\lambda }_{1}},{{\lambda }_{2}},{{\lambda }_{3}}......{{\lambda }_{n}}$ are constants in function $'y'$ as written below:

$y={{\lambda }_{1}}{{f}_{1}}(x)+{{\lambda }_{2}}{{f}_{2}}(x)+{{\lambda }_{3}}{{f}_{3}}(x)+......+{{\lambda }_{n}}{{f}_{n}}(x)$.

Now, if we differentiate the above given function $y$ , then we will get

$\begin{align}

& \dfrac{dy}{dx}=\dfrac{d}{dx}\left[ {{\lambda }_{1}}{{f}_{1}}(x)+{{\lambda }_{2}}{{f}_{2}}(x)+{{\lambda }_{3}}{{f}_{3}}(x)+......+{{\lambda }_{n}}{{f}_{n}}(x) \right] \\

& \dfrac{dy}{dx}=\dfrac{d}{dx}\left[ {{\lambda }_{1}}{{f}_{1}}(x) \right]+\dfrac{d}{dx}\left[ {{\lambda }_{2}}{{f}_{2}}(x) \right]+\dfrac{d}{dx}\left[ {{\lambda }_{3}}{{f}_{3}}(x) \right]+....+\dfrac{d}{dx}\left[ {{\lambda }_{n}}{{f}_{n}}(x) \right] \\

& \dfrac{dy}{dx}={{\lambda }_{1}}\dfrac{d}{dx}\left[ {{f}_{1}}(x) \right]+{{\lambda }_{2}}\dfrac{d}{dx}\left[ {{f}_{2}}(x) \right]+{{\lambda }_{3}}\dfrac{d}{dx}\left[ {{f}_{3}}(x) \right]+....+{{\lambda }_{n}}\dfrac{d}{dx}\left[ {{f}_{n}}(x) \right]........(2) \\

\end{align}$

So if functions are written in summation, then we can differentiate them individually. Using the above property in equation (1) as follows:

\[\begin{align}

& \dfrac{dy}{dx}=\dfrac{d}{dx}({{2}^{x}}-3{{e}^{x}}-{{4}^{x}}) \\

& \dfrac{d}{dx}=\dfrac{d}{dx}({{2}^{x}})-\dfrac{d}{dx}(3{{e}^{x}})-\dfrac{d}{dx}({{4}^{x}}) \\

\end{align}\]

Here, we can observe that ${{2}^{x}}$ is of type (Constant)function or ${{(\lambda )}^{f(x)}}$\ . So, we have formula for it as,

$\dfrac{d}{dx}({{a}^{x}})={{a}^{x}}\ln a$

Hence,

$\dfrac{d}{dx}({{2}^{x}})={{2}^{x}}\ln 2.........(4)$

Now, ${{e}^{x}}$ is exponential function and derivative of it is given as,

$\dfrac{d}{dx}({{e}^{x}})={{e}^{x}}........(5)$

Similarly, $\dfrac{d}{dx}({{4}^{x}})$ can be calculated by $\dfrac{d}{dx}({{a}^{x}})={{a}^{x}}\ln a$ formula as,

$\dfrac{d}{dx}({{4}^{x}})={{4}^{x}}\ln 4.........(6)$

Putting the values of equation (4), (5) and (6) in equation (3) for calculating,

\[\dfrac{dy}{dx}={{2}^{x}}\ln 2-3{{e}^{x}}-{{4}^{x}}\ln 4\]

Note: Student need to very clear with the functions like ${{a}^{x}},{{x}^{a}},{{x}^{x}},{{a}^{a}}$ .

Now we can calculate the differentiation of above by applying the multiplication rule.

Therefore, we need to be very clear with the above discussed functions and their differentiation.

So, we need to be very clear with the basic formulas.

The common mistake is student forget to notice (Constant)function or ${{(\lambda )}^{f(x)}}$kind of function and makes mistake in correct calculation, i.e.,

$\dfrac{d}{dx}({{a}^{x}})={{a}^{x}}\ln a$

We have expression/function

\[y={{2}^{x}}-3{{e}^{x}}-{{4}^{x}}-(1)\]

Now, let us differentiate the given function as,

\[\dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{2}^{x}}-3{{e}^{x}}-{{4}^{x}} \right)\]

One important rule should be applied here as stated below:

If we have $n$ functions ${{f}_{1}}(x),{{f}_{2}}(x),{{f}_{3}}(x)......{{f}_{n}}(x)$ and ${{\lambda }_{1}},{{\lambda }_{2}},{{\lambda }_{3}}......{{\lambda }_{n}}$ are constants in function $'y'$ as written below:

$y={{\lambda }_{1}}{{f}_{1}}(x)+{{\lambda }_{2}}{{f}_{2}}(x)+{{\lambda }_{3}}{{f}_{3}}(x)+......+{{\lambda }_{n}}{{f}_{n}}(x)$.

Now, if we differentiate the above given function $y$ , then we will get

$\begin{align}

& \dfrac{dy}{dx}=\dfrac{d}{dx}\left[ {{\lambda }_{1}}{{f}_{1}}(x)+{{\lambda }_{2}}{{f}_{2}}(x)+{{\lambda }_{3}}{{f}_{3}}(x)+......+{{\lambda }_{n}}{{f}_{n}}(x) \right] \\

& \dfrac{dy}{dx}=\dfrac{d}{dx}\left[ {{\lambda }_{1}}{{f}_{1}}(x) \right]+\dfrac{d}{dx}\left[ {{\lambda }_{2}}{{f}_{2}}(x) \right]+\dfrac{d}{dx}\left[ {{\lambda }_{3}}{{f}_{3}}(x) \right]+....+\dfrac{d}{dx}\left[ {{\lambda }_{n}}{{f}_{n}}(x) \right] \\

& \dfrac{dy}{dx}={{\lambda }_{1}}\dfrac{d}{dx}\left[ {{f}_{1}}(x) \right]+{{\lambda }_{2}}\dfrac{d}{dx}\left[ {{f}_{2}}(x) \right]+{{\lambda }_{3}}\dfrac{d}{dx}\left[ {{f}_{3}}(x) \right]+....+{{\lambda }_{n}}\dfrac{d}{dx}\left[ {{f}_{n}}(x) \right]........(2) \\

\end{align}$

So if functions are written in summation, then we can differentiate them individually. Using the above property in equation (1) as follows:

\[\begin{align}

& \dfrac{dy}{dx}=\dfrac{d}{dx}({{2}^{x}}-3{{e}^{x}}-{{4}^{x}}) \\

& \dfrac{d}{dx}=\dfrac{d}{dx}({{2}^{x}})-\dfrac{d}{dx}(3{{e}^{x}})-\dfrac{d}{dx}({{4}^{x}}) \\

\end{align}\]

Here, we can observe that ${{2}^{x}}$ is of type (Constant)function or ${{(\lambda )}^{f(x)}}$\ . So, we have formula for it as,

$\dfrac{d}{dx}({{a}^{x}})={{a}^{x}}\ln a$

Hence,

$\dfrac{d}{dx}({{2}^{x}})={{2}^{x}}\ln 2.........(4)$

Now, ${{e}^{x}}$ is exponential function and derivative of it is given as,

$\dfrac{d}{dx}({{e}^{x}})={{e}^{x}}........(5)$

Similarly, $\dfrac{d}{dx}({{4}^{x}})$ can be calculated by $\dfrac{d}{dx}({{a}^{x}})={{a}^{x}}\ln a$ formula as,

$\dfrac{d}{dx}({{4}^{x}})={{4}^{x}}\ln 4.........(6)$

Putting the values of equation (4), (5) and (6) in equation (3) for calculating,

\[\dfrac{dy}{dx}={{2}^{x}}\ln 2-3{{e}^{x}}-{{4}^{x}}\ln 4\]

Note: Student need to very clear with the functions like ${{a}^{x}},{{x}^{a}},{{x}^{x}},{{a}^{a}}$ .

Now we can calculate the differentiation of above by applying the multiplication rule.

Therefore, we need to be very clear with the above discussed functions and their differentiation.

So, we need to be very clear with the basic formulas.

The common mistake is student forget to notice (Constant)function or ${{(\lambda )}^{f(x)}}$kind of function and makes mistake in correct calculation, i.e.,

$\dfrac{d}{dx}({{a}^{x}})={{a}^{x}}\ln a$

Last updated date: 29th Sep 2023

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Total views: 364.2k

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