Solved on Dec 27, 2023

Differentiate $e^{3x}(\sin x + 2\cos x)$ with respect to $x$ .

STEP 1

Assumptions
1. We are differentiating the function $f(x) = e^{3x}(\sin x + 2\cos x)$ with respect to $x$ .
2. We will apply the product rule for differentiation, which states that if $u(x)$ and $v(x)$ are differentiable functions of $x$ , then the derivative of $u(x)v(x)$ is given by $u'(x)v(x) + u(x)v'(x)$ .
3. We will also apply the chain rule for differentiation, which states that if $y = g(u)$ and $u = f(x)$ are differentiable functions, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ .
4. We will use the basic derivatives that $\frac{d}{dx}e^{ax} = ae^{ax}$ , $\frac{d}{dx}\sin x = \cos x$ , and $\frac{d}{dx}\cos x = -\sin x$ .

STEP 2

Identify the two functions that are being multiplied in the given function.
Let $u(x) = e^{3x}$ and $v(x) = \sin x + 2\cos x$ .

STEP 3

Differentiate $u(x)$ with respect to $x$ using the chain rule.
$u'(x) = \frac{d}{dx}e^{3x} = 3e^{3x}$

STEP 4

Differentiate $v(x)$ with respect to $x$ using the sum rule and the basic derivatives of $\sin x$ and $\cos x$ .
$v'(x) = \frac{d}{dx}(\sin x + 2\cos x) = \cos x - 2\sin x$

STEP 5

Apply the product rule to differentiate $f(x) = u(x)v(x)$ .
$f'(x) = u'(x)v(x) + u(x)v'(x)$

STEP 6

Substitute $u(x)$ , $v(x)$ , $u'(x)$ , and $v'(x)$ into the product rule formula.
$f'(x) = (3e^{3x})(\sin x + 2\cos x) + (e^{3x})(\cos x - 2\sin x)$

STEP 7

Distribute $3e^{3x}$ through the first set of parentheses and $e^{3x}$ through the second set of parentheses.
$f'(x) = 3e^{3x}\sin x + 6e^{3x}\cos x + e^{3x}\cos x - 2e^{3x}\sin x$

STEP 8

Combine like terms.
$f'(x) = (3e^{3x}\sin x - 2e^{3x}\sin x) + (6e^{3x}\cos x + e^{3x}\cos x)$

STEP 9

Simplify the expression by combining the coefficients of like terms.
$f'(x) = e^{3x}\sin x + 7e^{3x}\cos x$

STEP 10

Write the final simplified expression for the derivative of the given function.
$\frac{d}{dx}e^{3x}(\sin x + 2\cos x) = e^{3x}\sin x + 7e^{3x}\cos x$
The derivative with respect to $x$ of the function $e^{3x}(\sin x + 2\cos x)$ is $e^{3x}\sin x + 7e^{3x}\cos x$ .

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Differentiate $e^{3x}(\sin x + 2\cos x)$ with respect to $x$ .

STEP 1

STEP 2

Identify the two functions that are being multiplied in the given function.
Let $u(x) = e^{3x}$ and $v(x) = \sin x + 2\cos x$ .

STEP 3

Differentiate $u(x)$ with respect to $x$ using the chain rule.
$u'(x) = \frac{d}{dx}e^{3x} = 3e^{3x}$

STEP 4

Differentiate $v(x)$ with respect to $x$ using the sum rule and the basic derivatives of $\sin x$ and $\cos x$ .
$v'(x) = \frac{d}{dx}(\sin x + 2\cos x) = \cos x - 2\sin x$

STEP 5

Apply the product rule to differentiate $f(x) = u(x)v(x)$ .
$f'(x) = u'(x)v(x) + u(x)v'(x)$

STEP 6

Substitute $u(x)$ , $v(x)$ , $u'(x)$ , and $v'(x)$ into the product rule formula.
$f'(x) = (3e^{3x})(\sin x + 2\cos x) + (e^{3x})(\cos x - 2\sin x)$

STEP 7

Distribute $3e^{3x}$ through the first set of parentheses and $e^{3x}$ through the second set of parentheses.
$f'(x) = 3e^{3x}\sin x + 6e^{3x}\cos x + e^{3x}\cos x - 2e^{3x}\sin x$

STEP 8

Combine like terms.
$f'(x) = (3e^{3x}\sin x - 2e^{3x}\sin x) + (6e^{3x}\cos x + e^{3x}\cos x)$

STEP 9

Simplify the expression by combining the coefficients of like terms.
$f'(x) = e^{3x}\sin x + 7e^{3x}\cos x$

STEP 10

Write the final simplified expression for the derivative of the given function.
$\frac{d}{dx}e^{3x}(\sin x + 2\cos x) = e^{3x}\sin x + 7e^{3x}\cos x$
The derivative with respect to $x$ of the function $e^{3x}(\sin x + 2\cos x)$ is $e^{3x}\sin x + 7e^{3x}\cos x$ .

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Differentiate e3x(sin⁡x+2cos⁡x)e^{3x}(\sin x + 2\cos x)e3x(sinx+2cosx) with respect to xxx.

STEP 1

STEP 2

Identify the two functions that are being multiplied in the given function.Let u(x)=e3xu(x) = e^{3x}u(x)=e3x and v(x)=sin⁡x+2cos⁡xv(x) = \sin x + 2\cos xv(x)=sinx+2cosx.

STEP 3

Differentiate u(x)u(x)u(x) with respect to xxx using the chain rule.u′(x)=ddxe3x=3e3xu'(x) = \frac{d}{dx}e^{3x} = 3e^{3x}u′(x)=dxd​e3x=3e3x

STEP 4

Differentiate v(x)v(x)v(x) with respect to xxx using the sum rule and the basic derivatives of sin⁡x\sin xsinx and cos⁡x\cos xcosx.v′(x)=ddx(sin⁡x+2cos⁡x)=cos⁡x−2sin⁡xv'(x) = \frac{d}{dx}(\sin x + 2\cos x) = \cos x - 2\sin xv′(x)=dxd​(sinx+2cosx)=cosx−2sinx

STEP 5

Apply the product rule to differentiate f(x)=u(x)v(x)f(x) = u(x)v(x)f(x)=u(x)v(x).f′(x)=u′(x)v(x)+u(x)v′(x)f'(x) = u'(x)v(x) + u(x)v'(x)f′(x)=u′(x)v(x)+u(x)v′(x)

STEP 6

Substitute u(x)u(x)u(x), v(x)v(x)v(x), u′(x)u'(x)u′(x), and v′(x)v'(x)v′(x) into the product rule formula.f′(x)=(3e3x)(sin⁡x+2cos⁡x)+(e3x)(cos⁡x−2sin⁡x)f'(x) = (3e^{3x})(\sin x + 2\cos x) + (e^{3x})(\cos x - 2\sin x)f′(x)=(3e3x)(sinx+2cosx)+(e3x)(cosx−2sinx)

STEP 7

Distribute 3e3x3e^{3x}3e3x through the first set of parentheses and e3xe^{3x}e3x through the second set of parentheses.f′(x)=3e3xsin⁡x+6e3xcos⁡x+e3xcos⁡x−2e3xsin⁡xf'(x) = 3e^{3x}\sin x + 6e^{3x}\cos x + e^{3x}\cos x - 2e^{3x}\sin xf′(x)=3e3xsinx+6e3xcosx+e3xcosx−2e3xsinx

STEP 8

Combine like terms.f′(x)=(3e3xsin⁡x−2e3xsin⁡x)+(6e3xcos⁡x+e3xcos⁡x)f'(x) = (3e^{3x}\sin x - 2e^{3x}\sin x) + (6e^{3x}\cos x + e^{3x}\cos x)f′(x)=(3e3xsinx−2e3xsinx)+(6e3xcosx+e3xcosx)

STEP 9

Simplify the expression by combining the coefficients of like terms.f′(x)=e3xsin⁡x+7e3xcos⁡xf'(x) = e^{3x}\sin x + 7e^{3x}\cos xf′(x)=e3xsinx+7e3xcosx

STEP 10

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Differentiate $e^{3x}(\sin x + 2\cos x)$ with respect to $x$ .

Identify the two functions that are being multiplied in the given function.
Let $u(x) = e^{3x}$ and $v(x) = \sin x + 2\cos x$ .

Differentiate $u(x)$ with respect to $x$ using the chain rule.
$u'(x) = \frac{d}{dx}e^{3x} = 3e^{3x}$

Differentiate $v(x)$ with respect to $x$ using the sum rule and the basic derivatives of $\sin x$ and $\cos x$ .
$v'(x) = \frac{d}{dx}(\sin x + 2\cos x) = \cos x - 2\sin x$

Apply the product rule to differentiate $f(x) = u(x)v(x)$ .
$f'(x) = u'(x)v(x) + u(x)v'(x)$

Substitute $u(x)$ , $v(x)$ , $u'(x)$ , and $v'(x)$ into the product rule formula.
$f'(x) = (3e^{3x})(\sin x + 2\cos x) + (e^{3x})(\cos x - 2\sin x)$

Distribute $3e^{3x}$ through the first set of parentheses and $e^{3x}$ through the second set of parentheses.
$f'(x) = 3e^{3x}\sin x + 6e^{3x}\cos x + e^{3x}\cos x - 2e^{3x}\sin x$

Combine like terms.
$f'(x) = (3e^{3x}\sin x - 2e^{3x}\sin x) + (6e^{3x}\cos x + e^{3x}\cos x)$

Simplify the expression by combining the coefficients of like terms.
$f'(x) = e^{3x}\sin x + 7e^{3x}\cos x$