Question

How do you differentiate $y = \left( {{e^{ - 4x}}} \right)$ with respect to $x$?

Answer 1

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.