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Hapa kuna suluhisho la maswali yako ya calculus: a) Differentiate y = 3x² − 7x from first principles. Ili kutofautisha kutoka kanuni za kwanza, tunatumia fomula: f'(x) = lim_(h → 0) (f(x+h) − f(x))/(h); Hapa, f(x) = 3x² − 7x. Step 1: Tafuta f(x+h). f(x+h) = 3(x+h)² − 7(x+h); f(x+h) = 3(x² + 2xh + h²) − 7x − 7h; f(x+h) = 3x² + 6xh + 3h² − 7x − 7h Step 2: Tafuta f(x+h) − f(x). f(x+h) − f(x) = (3x² + 6xh + 3h² − 7x − 7h) − (3x² − 7x); f(x+h) − f(x) = 3x² + 6xh + 3h² − 7x − 7h − 3x² + 7x; f(x+h) − f(x) = 6xh + 3h² − 7h Step 3: Gawanya kwa h. (f(x+h) − f(x))/(h) = (6xh + 3h² − 7h)/(h); (f(x+h) − f(x))/(h) = 6x + 3h − 7 Step 4: Chukua kikomo kama h → 0. f'(x) = lim_(h → 0) (6x + 3h − 7); f'(x) = 6x − 7 Jibu la mwisho: dy/dx = 6x − 7 b) Differentiate each of the following functions: (i) y = (x−9)⁷ Tunatumia sheria ya mnyororo (chain rule). Step 1: Acha u = x−9. Kisha y = u⁷. Step 2: Tafuta dy/du na du/dx. dy/du = 7u⁶; du/dx = 1; Step 3: Tumia sheria ya mnyororo dy/dx = dy/du · du/dx. dy/dx = 7u⁶ · 1; dy/dx = 7(x−9)⁶ Jibu la mwisho: dy/dx = 7(x−9)⁶; (ii) y = x² sin 3x; Tunatumia sheria ya bidhaa (product rule): d/dx(uv) = u'v + uv'. Step 1: Acha u = x² na v = sin 3x. Step 2: Tafuta u' na v'. u' = d/dx(x²) = 2x; v' = d/dx(sin 3x) = cos 3x · d/dx(3x) = 3 cos 3x Step 3: Tumia sheria ya bidhaa. dy/dx = (2x)(sin 3x) + (x²)(3 cos 3x); dy/dx = 2x sin 3x + 3x² cos 3x Jibu la mwisho: dy/dx = 2x sin 3x + 3x² cos 3x; (iii) y = (x³+1)/(x+1); Kwanza, rahisiisha usemi kwa kutumia utambulisho wa jumla ya cubes a³+b³ = (a+b)(a²−ab+b²). Step 1: Rahisiisha usemi. y = ((x+1)(x²−x+1))/(x+1) Kwa x ≠ −1, usemi unakuwa: y = x²−x+1 Step 2: Tofautisha usemi uliorahisishwa. dy/dx = d/dx(x²−x+1); dy/dx = 2x − 1 Jibu la mwisho: dy/dx = 2x − 1; c) Locate the stationary points of the function z = x³ + y³ − 3x − 6y − 1 and hence determine their nature. Ili kupata pointi za stationary, tunapata derivatives za sehemu ya kwanza na kuziweka sawa na sifuri. Step 1: Tafuta derivatives za sehemu ya kwanza. (∂ z)/(∂ x) = 3x² − 3; (∂ z)/(∂ y) = 3y² − 6 Step 2: Weka derivatives za sehemu sawa na sifuri na utatue kwa x na y. 3x² − 3 = 0 ⇒ 3x² = 3 ⇒ x² = 1 ⇒ x = ± 1; 3y² − 6 = 0 ⇒ 3y² = 6 ⇒ y² = 2 ⇒ y = ± √(2) Pointi za stationary ni: (1, √(2)), (1, −√(2)), (−1, √(2)), (−1, −√(2)). Step 3: Tafuta derivatives za sehemu ya pili ili kuamua asili ya pointi. (∂² z)/(∂ x²) = 6x; (∂² z)/(∂ y²) = 6y; (∂² z)/(∂ x ∂ y) = 0; Tunatumia mtihani wa derivative ya pili na discriminant D = (∂² z)/(∂ x²) (∂² z)/(∂ y²) − ((∂² z)/(∂ x ∂ y))². Step 4: Tathmini kila pointi ya stationary. Kwa pointi (1, √(2)): (∂² z)/(∂ x²) = 6(1) = 6; (∂² z)/(∂ y²) = 6√(2); D = (6)(6√(2)) − (0)² = 36√(2) Kwa kuwa D > 0 na (∂² z)/(∂ x²) > 0, pointi hii ni local minimum. Kwa pointi (1, −√(2)): (∂² z)/(∂ x²) = 6(1) = 6; (∂² z)/(∂ y²) = 6(−√(2)) = −6√(2); D = (6)(−6√(2)) − (0)² = −36√(2) Kwa kuwa D < 0, pointi hii ni saddle point. Kwa pointi (−1, √(2)): (∂² z)/(∂ x²) = 6(−1) = −6; (∂² z)/(∂ y²) = 6√(2); D = (−6)(6√(2)) − (0)² = −36√(2) Kwa kuwa D < 0, pointi hii ni saddle point. Kwa pointi (−1, −√(2)): (∂² z)/(∂ x²) = 6(−1) = −6; (∂² z)/(∂ y²) = 6(−√(2)) = −6√(2); D = (−6)(−6√(2)) − (0)² = 36√(2) Kwa kuwa D > 0 na (∂² z)/(∂ x²) < 0, pointi hii ni local maximum. Jibu la mwisho: Pointi za stationary na asili yake ni: (1, √(2)): local minimum (1, −√(2)): saddle point (−1, √(2)): saddle point (−1, −√(2)): local maximum

Give calculation in English

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Step-by-step solution

Part a· 4 steps

Tafuta $f(x+h)$ .

Tafuta $f(x+h) - f(x)$ .

Gawanya kwa $h$ .

Chukua kikomo kama $h \to 0$ .

Part b(i)· 3 steps
$\frac{dy}{dx} = 7(x-9)^6$

Acha $u = x-9$ .

Tafuta $\frac{dy}{du}$ na $\frac{du}{dx}$ .

Tumia sheria ya mnyororo $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ .

Part b(ii)· 3 steps
$\frac{dy}{dx} = 2x \sin 3x + 3x^2 \cos 3x$

Acha $u = x^2$ na $v = \sin 3x$ .

Tafuta $u'$ na $v'$ .

Tumia sheria ya bidhaa.

Part b(iii)· 2 steps
$\frac{dy}{dx} = 2x - 1$

Rahisiisha usemi.

Tofautisha usemi uliorahisishwa.

Part c· 4 steps

Tafuta derivatives za sehemu ya kwanza.

Weka derivatives za sehemu sawa na sifuri na utatue kwa $x$ na $y$ .

Tafuta derivatives za sehemu ya pili ili kuamua asili ya pointi.

Tathmini kila pointi ya stationary.

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Give calculation in English

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Step-by-step solution

Part a· 4 steps

Tafuta f(x+h)f(x+h)f(x+h).

Tafuta f(x+h)−f(x)f(x+h) - f(x)f(x+h)−f(x).

Gawanya kwa hhh.

Chukua kikomo kama h→0h \to 0h→0.

Part b(i)· 3 stepsdydx=7(x−9)6\frac{dy}{dx} = 7(x-9)^6dxdy​=7(x−9)6

Acha u=x−9u = x-9u=x−9.

Tafuta dydu\frac{dy}{du}dudy​ na dudx\frac{du}{dx}dxdu​.

Tumia sheria ya mnyororo dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}dxdy​=dudy​⋅dxdu​.

Part b(ii)· 3 stepsdydx=2xsin⁡3x+3x2cos⁡3x\frac{dy}{dx} = 2x \sin 3x + 3x^2 \cos 3xdxdy​=2xsin3x+3x2cos3x

Acha u=x2u = x^2u=x2 na v=sin⁡3xv = \sin 3xv=sin3x.

Tafuta u′u'u′ na v′v'v′.

Tumia sheria ya bidhaa.

Part b(iii)· 2 stepsdydx=2x−1\frac{dy}{dx} = 2x - 1dxdy​=2x−1

Rahisiisha usemi.

Tofautisha usemi uliorahisishwa.

Part c· 4 steps

Tafuta derivatives za sehemu ya kwanza.

Weka derivatives za sehemu sawa na sifuri na utatue kwa xxx na yyy.

Tafuta derivatives za sehemu ya pili ili kuamua asili ya pointi.

Tathmini kila pointi ya stationary.

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Still stuck on this one?

More English Questions

Tafuta $f(x+h)$ .

Tafuta $f(x+h) - f(x)$ .

Gawanya kwa $h$ .

Chukua kikomo kama $h \to 0$ .

Part b(i)· 3 steps
$\frac{dy}{dx} = 7(x-9)^6$

Acha $u = x-9$ .

Tafuta $\frac{dy}{du}$ na $\frac{du}{dx}$ .

Tumia sheria ya mnyororo $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ .

Part b(ii)· 3 steps
$\frac{dy}{dx} = 2x \sin 3x + 3x^2 \cos 3x$

Acha $u = x^2$ na $v = \sin 3x$ .

Tafuta $u'$ na $v'$ .

Part b(iii)· 2 steps
$\frac{dy}{dx} = 2x - 1$

Weka derivatives za sehemu sawa na sifuri na utatue kwa $x$ na $y$ .