How do you differentiate $y = \left( {{e^{ - 4x}}} \right)$ with respect to $x$?
Answer
623.1k+ views
Hint:In the given problem, we are required to differentiate $y = \left( {{e^{ - 4x}}} \right)$ with respect to x. Since, $y = \left( {{e^{ - 4x}}} \right)$ is a composite function, so we will have to apply chain rule of differentiation in the process of differentiating $y = \left( {{e^{ - 4x}}} \right)$ . So, differentiation of $y = \left( {{e^{ - 4x}}} \right)$ with respect to x will be done layer by layer using the chain rule of differentiation. Also the derivative of $y = \left( {{e^x}} \right)$with respect to x must be remembered.
Complete step by step answer:
To find derivative of $y = \left( {{e^{ - 4x}}} \right)$ with respect to x we have to find differentiate $y = \left( {{e^{ - 4x}}} \right)$with respect to x.
So, Derivative of $y = \left( {{e^{ - 4x}}} \right)$ with respect to x can be calculated as $\dfrac{d}{{dx}}\left( {{e^{ - 4x}}} \right)$ .
Now, $\dfrac{d}{{dx}}\left( {{e^{ - 4x}}} \right)$
So, first differentiating ${e^{ - 4x}}$ with respect to $\left( { - 4x} \right)$, we get, ${e^{ - 4x}}$and then differentiate $ - 4x$ with respect to x and get $ - 4$ as the derivative.
\[\dfrac{d}{{dx}}\left[ {{e^{ - 4x}}} \right]\]
Now, Let us assume $u = - 4x$. So substituting $ - 4x$as $u$, we get,
$\dfrac{d}{{dx}}\left( {{e^u}} \right)$
$\Rightarrow\;{e^u}\dfrac{{du}}{{dx}}$
Now, putting back $u$as $ - 4x$, we get,
$\;{e^{\left( { - 4x} \right)}}\dfrac{d}{{dx}}\left( { - 4x} \right)$
$\therefore - 4\;{e^{ - 4x}}$
So, the derivative of ${e^{ - 4x}}$ with respect to $x$is $ - 4\;{e^{ - 4x}}$.
Note:The given problem may also be solved using the first principle of differentiation. The derivatives of basic trigonometric functions must be learned by heart in order to find derivatives of complex composite functions using chain rule of differentiation. The chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer.
Complete step by step answer:
To find derivative of $y = \left( {{e^{ - 4x}}} \right)$ with respect to x we have to find differentiate $y = \left( {{e^{ - 4x}}} \right)$with respect to x.
So, Derivative of $y = \left( {{e^{ - 4x}}} \right)$ with respect to x can be calculated as $\dfrac{d}{{dx}}\left( {{e^{ - 4x}}} \right)$ .
Now, $\dfrac{d}{{dx}}\left( {{e^{ - 4x}}} \right)$
So, first differentiating ${e^{ - 4x}}$ with respect to $\left( { - 4x} \right)$, we get, ${e^{ - 4x}}$and then differentiate $ - 4x$ with respect to x and get $ - 4$ as the derivative.
\[\dfrac{d}{{dx}}\left[ {{e^{ - 4x}}} \right]\]
Now, Let us assume $u = - 4x$. So substituting $ - 4x$as $u$, we get,
$\dfrac{d}{{dx}}\left( {{e^u}} \right)$
$\Rightarrow\;{e^u}\dfrac{{du}}{{dx}}$
Now, putting back $u$as $ - 4x$, we get,
$\;{e^{\left( { - 4x} \right)}}\dfrac{d}{{dx}}\left( { - 4x} \right)$
$\therefore - 4\;{e^{ - 4x}}$
So, the derivative of ${e^{ - 4x}}$ with respect to $x$is $ - 4\;{e^{ - 4x}}$.
Note:The given problem may also be solved using the first principle of differentiation. The derivatives of basic trigonometric functions must be learned by heart in order to find derivatives of complex composite functions using chain rule of differentiation. The chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer.
Recently Updated Pages
What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

10 examples of friction in our daily life

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

A Paragraph on Pollution in about 100-150 Words

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

