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Answer :
We are given the function
[tex]$$
y = x e^{-3x} + 7 e^{-x} + x^7.
$$[/tex]
We will differentiate this function term by term.
1. First, consider the term
[tex]$$
y_1 = x e^{-3x}.
$$[/tex]
This term is a product of two functions: [tex]$u(x) = x$[/tex] and [tex]$v(x) = e^{-3x}$[/tex]. By the product rule,
[tex]$$
\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x).
$$[/tex]
Calculate the derivatives:
- [tex]$u'(x) = \frac{d}{dx}(x) = 1$[/tex].
- To find [tex]$v'(x)$[/tex], note that [tex]$v(x) = e^{-3x}$[/tex]. Differentiating gives
[tex]$$
v'(x) = \frac{d}{dx}\left(e^{-3x}\right) = -3 e^{-3x}.
$$[/tex]
So, applying the product rule:
[tex]$$
\frac{d}{dx}\left(x e^{-3x}\right) = 1\cdot e^{-3x} + x\cdot (-3e^{-3x}) = e^{-3x} - 3x e^{-3x}.
$$[/tex]
2. Next, consider the term
[tex]$$
y_2 = 7 e^{-x}.
$$[/tex]
Since [tex]$7$[/tex] is a constant and [tex]$e^{-x}$[/tex] is an exponential function, we apply the chain rule. The derivative of [tex]$e^{-x}$[/tex] is
[tex]$$
\frac{d}{dx}(e^{-x}) = -e^{-x},
$$[/tex]
so
[tex]$$
\frac{d}{dx}\left(7 e^{-x}\right) = 7\cdot(-e^{-x}) = -7 e^{-x}.
$$[/tex]
3. Finally, for the term
[tex]$$
y_3 = x^7,
$$[/tex]
we use the power rule:
[tex]$$
\frac{d}{dx}(x^7) = 7 x^6.
$$[/tex]
Now, adding the derivatives of each term, we get the overall derivative:
[tex]$$
\frac{dy}{dx} = \left(e^{-3x} - 3x e^{-3x}\right) + \left(-7 e^{-x}\right) + 7 x^6.
$$[/tex]
This simplifies to:
[tex]$$
\frac{dy}{dx} = e^{-3x} - 3x e^{-3x} - 7 e^{-x} + 7 x^6.
$$[/tex]
Among the choices provided, this corresponds to option D:
[tex]$$
\frac{dy}{dx} = -3 x e^{-3x} + e^{-3x} - 7 e^{-x} + 7 x^6.
$$[/tex]
Thus, the correct answer is option D.
[tex]$$
y = x e^{-3x} + 7 e^{-x} + x^7.
$$[/tex]
We will differentiate this function term by term.
1. First, consider the term
[tex]$$
y_1 = x e^{-3x}.
$$[/tex]
This term is a product of two functions: [tex]$u(x) = x$[/tex] and [tex]$v(x) = e^{-3x}$[/tex]. By the product rule,
[tex]$$
\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x).
$$[/tex]
Calculate the derivatives:
- [tex]$u'(x) = \frac{d}{dx}(x) = 1$[/tex].
- To find [tex]$v'(x)$[/tex], note that [tex]$v(x) = e^{-3x}$[/tex]. Differentiating gives
[tex]$$
v'(x) = \frac{d}{dx}\left(e^{-3x}\right) = -3 e^{-3x}.
$$[/tex]
So, applying the product rule:
[tex]$$
\frac{d}{dx}\left(x e^{-3x}\right) = 1\cdot e^{-3x} + x\cdot (-3e^{-3x}) = e^{-3x} - 3x e^{-3x}.
$$[/tex]
2. Next, consider the term
[tex]$$
y_2 = 7 e^{-x}.
$$[/tex]
Since [tex]$7$[/tex] is a constant and [tex]$e^{-x}$[/tex] is an exponential function, we apply the chain rule. The derivative of [tex]$e^{-x}$[/tex] is
[tex]$$
\frac{d}{dx}(e^{-x}) = -e^{-x},
$$[/tex]
so
[tex]$$
\frac{d}{dx}\left(7 e^{-x}\right) = 7\cdot(-e^{-x}) = -7 e^{-x}.
$$[/tex]
3. Finally, for the term
[tex]$$
y_3 = x^7,
$$[/tex]
we use the power rule:
[tex]$$
\frac{d}{dx}(x^7) = 7 x^6.
$$[/tex]
Now, adding the derivatives of each term, we get the overall derivative:
[tex]$$
\frac{dy}{dx} = \left(e^{-3x} - 3x e^{-3x}\right) + \left(-7 e^{-x}\right) + 7 x^6.
$$[/tex]
This simplifies to:
[tex]$$
\frac{dy}{dx} = e^{-3x} - 3x e^{-3x} - 7 e^{-x} + 7 x^6.
$$[/tex]
Among the choices provided, this corresponds to option D:
[tex]$$
\frac{dy}{dx} = -3 x e^{-3x} + e^{-3x} - 7 e^{-x} + 7 x^6.
$$[/tex]
Thus, the correct answer is option D.
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