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1. Cari turunan dari 3 x − 5 x 2 + 7 \frac{3x - 5}{x^2 + 7} x 2 + 7 3 x − 5
This is a derivative of a fraction using the quotient rule:
u v ′ = u ′ v − u v ′ v 2 \frac{u}{v}' = \frac{u'v - uv'}{v^2} v u ′ = v 2 u ′ v − u v ′
Here, u = 3 x − 5 u = 3x - 5 u = 3 x − 5 and v = x 2 + 7 v = x^2 + 7 v = x 2 + 7 .
u ′ = 3 u' = 3 u ′ = 3 , v ′ = 2 x v' = 2x v ′ = 2 x .
Substituting:
d d x ( 3 x − 5 x 2 + 7 ) = ( 3 ) ( x 2 + 7 ) − ( 3 x − 5 ) ( 2 x ) ( x 2 + 7 ) 2 \frac{d}{dx}\left(\frac{3x - 5}{x^2 + 7}\right) = \frac{(3)(x^2 + 7) - (3x - 5)(2x)}{(x^2 + 7)^2} d x d ( x 2 + 7 3 x − 5 ) = ( x 2 + 7 ) 2 ( 3 ) ( x 2 + 7 ) − ( 3 x − 5 ) ( 2 x )
Simplify the numerator:
3 ( x 2 + 7 ) − ( 3 x − 5 ) ( 2 x ) = 3 x 2 + 21 − 6 x 2 + 10 x = − 3 x 2 + 10 x + 21 3(x^2 + 7) - (3x - 5)(2x) = 3x^2 + 21 - 6x^2 + 10x = -3x^2 + 10x + 21 3 ( x 2 + 7 ) − ( 3 x − 5 ) ( 2 x ) = 3 x 2 + 21 − 6 x 2 + 10 x = − 3 x 2 + 10 x + 21
Final result:
− 3 x 2 + 10 x + 21 ( x 2 + 7 ) 2 \frac{-3x^2 + 10x + 21}{(x^2 + 7)^2} ( x 2 + 7 ) 2 − 3 x 2 + 10 x + 21
2. Cari d y dy d y jika y = 2 x 4 + 1 + 3 x y = \frac{2}{x^4 + 1} + \frac{3}{x} y = x 4 + 1 2 + x 3
Here, y y y is the sum of two terms, so we find d y / d x dy/dx d y / d x term by term.
For 2 x 4 + 1 \frac{2}{x^4 + 1} x 4 + 1 2 :
Use the chain rule:
d d x ( 2 x 4 + 1 ) = − 2 ⋅ 4 x 3 ( x 4 + 1 ) 2 = − 8 x 3 ( x 4 + 1 ) 2 \frac{d}{dx}\left(\frac{2}{x^4 + 1}\right) = -\frac{2 \cdot 4x^3}{(x^4 + 1)^2} = -\frac{8x^3}{(x^4 + 1)^2} d x d ( x 4 + 1 2 ) = − ( x 4 + 1 ) 2 2 ⋅ 4 x 3 = − ( x 4 + 1 ) 2 8 x 3
For 3 x \frac{3}{x} x 3 :
This simplifies as:
d d x ( 3 x ) = − 3 x 2 \frac{d}{dx}\left(\frac{3}{x}\right) = -\frac{3}{x^2} d x d ( x 3 ) = − x 2 3
Combine the results:
d y = − 8 x 3 ( x 4 + 1 ) 2 − 3 x 2 dy = -\frac{8x^3}{(x^4 + 1)^2} - \frac{3}{x^2} d y = − ( x 4 + 1 ) 2 8 x 3 − x 2 3
3. Cari limit-limit berikut:
a. lim t → 0 1 − cos t sin t \lim_{t \to 0} \frac{1 - \cos t}{\sin t} lim t → 0 s i n t 1 − c o s t
Apply trigonometric identities:
Use 1 − cos t = 2 sin 2 ( t / 2 ) 1 - \cos t = 2\sin^2(t/2) 1 − cos t = 2 sin 2 ( t /2 ) :
lim t → 0 1 − cos t sin t = lim t → 0 2 sin 2 ( t / 2 ) sin t \lim_{t \to 0} \frac{1 - \cos t}{\sin t} = \lim_{t \to 0} \frac{2\sin^2(t/2)}{\sin t} lim t → 0 s i n t 1 − c o s t = lim t → 0 s i n t 2 s i n 2 ( t /2 )
Substituting sin ( t ) ≈ t \sin(t) \approx t sin ( t ) ≈ t for small t t t :
lim t → 0 2 sin 2 ( t / 2 ) sin t = 2 ( 1 / 2 ) 2 1 = 0 \lim_{t \to 0} \frac{2\sin^2(t/2)}{\sin t} = \frac{2(1/2)^2}{1} = 0 lim t → 0 s i n t 2 s i n 2 ( t /2 ) = 1 2 ( 1/2 ) 2 = 0
b. lim x → 0 sin 5 x 3 x \lim_{x \to 0} \frac{\sin 5x}{3x} lim x → 0 3 x s i n 5 x
Use the small angle approximation sin ( 5 x ) ≈ 5 x \sin(5x) \approx 5x sin ( 5 x ) ≈ 5 x :
lim x → 0 sin 5 x 3 x = lim x → 0 5 x 3 x = 5 3 \lim_{x \to 0} \frac{\sin 5x}{3x} = \lim_{x \to 0} \frac{5x}{3x} = \frac{5}{3} lim x → 0 3 x s i n 5 x = lim x → 0 3 x 5 x = 3 5
4. Cari d y / d x dy/dx d y / d x jika 4 x 2 y − 3 y = x 3 − 1 4x^2y - 3y = x^3 - 1 4 x 2 y − 3 y = x 3 − 1
Implicit differentiation:
Differentiate both sides:
d d x ( 4 x 2 y − 3 y ) = d d x ( x 3 − 1 ) \frac{d}{dx}(4x^2y - 3y) = \frac{d}{dx}(x^3 - 1) d x d ( 4 x 2 y − 3 y ) = d x d ( x 3 − 1 )
Product rule for 4 x 2 y 4x^2y 4 x 2 y :
4 x 2 d y d x + 8 x y − 3 d y d x = 3 x 2 4x^2 \frac{dy}{dx} + 8xy - 3\frac{dy}{dx} = 3x^2 4 x 2 d x d y + 8 x y − 3 d x d y = 3 x 2
Collect terms with d y d x \frac{dy}{dx} d x d y :
( 4 x 2 − 3 ) d y d x = 3 x 2 − 8 x y (4x^2 - 3)\frac{dy}{dx} = 3x^2 - 8xy ( 4 x 2 − 3 ) d x d y = 3 x 2 − 8 x y
Solve for d y d x \frac{dy}{dx} d x d y :
d y d x = 3 x 2 − 8 x y 4 x 2 − 3 \frac{dy}{dx} = \frac{3x^2 - 8xy}{4x^2 - 3} d x d y = 4 x 2 − 3 3 x 2 − 8 x y
5. Optimasi Pagar
Given 100 m of fencing to create two adjacent rectangular pens:
Total length of fencing: 2 l + 3 w = 100 2l + 3w = 100 2 l + 3 w = 100 .
Solve for l l l in terms of w w w : l = 100 − 3 w 2 l = \frac{100 - 3w}{2} l = 2 100 − 3 w .
Area A = l w = ( 100 − 3 w 2 ) w A = lw = \left(\frac{100 - 3w}{2}\right)w A = lw = ( 2 100 − 3 w ) w :
A = 50 w − 3 w 2 2 A = 50w - \frac{3w^2}{2} A = 50 w − 2 3 w 2
Differentiate to find maximum area:
d A d w = 50 − 3 w = 0 ⟹ w = 50 3 \frac{dA}{dw} = 50 - 3w = 0 \implies w = \frac{50}{3} d w d A = 50 − 3 w = 0 ⟹ w = 3 50
Substitute w w w back to find l l l :
l = 100 − 3 ( 50 3 ) 2 = 25 l = \frac{100 - 3\left(\frac{50}{3}\right)}{2} = 25 l = 2 100 − 3 ( 3 50 ) = 25
Dimensions: l = 25 l = 25 l = 25 , w = 50 3 w = \frac{50}{3} w = 3 50 .
6. Biaya Minimum
Cost function:
C ( v ) = 30 + v 2 v + 14 x v C(v) = \frac{30 + v^2}{v} + \frac{14x}{v} C ( v ) = v 30 + v 2 + v 14 x
Differentiate:
d C d v = − 30 + v 2 v 2 + 2 v v − 14 x v 2 \frac{dC}{dv} = -\frac{30 + v^2}{v^2} + \frac{2v}{v} - \frac{14x}{v^2} d v d C = − v 2 30 + v 2 + v 2 v − v 2 14 x
Solve for v v v within 40 ≤ v ≤ 60 40 \leq v \leq 60 40 ≤ v ≤ 60 . I recommend numerical methods to finalize this.
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Tip: Always verify your derivative by substitution or checking boundary conditions!