Find dy/dx for the following: y=x2+1, (b) y=(x2-1)/(2x+1), (c) y=sqrt(x)−1/sqrt(x), (d) y=3x/2x+1, (e)…

Question

Find dy/dx for the following: y=x2+1, (b) y=(x2-1)/(2x+1), (c) y=sqrt(x)−1/sqrt(x), (d) y=3x/2x+1, (e) y=x/(x+4)3, (f) y=sqrt(x2-1), (g) y=x2/sqrt(2x+1), (h) y=−1/(1+sqrt(x))2, (i) y=1/(x2+2x−4)

ScanSolve AI Tutor · Accepted Answer

Sawa Shari, nimeelewa. Nitatoa majibu mafupi na ya moja kwa moja.

Hebu tutatue tatizo (q).

Tatizo ni:

y = \frac{x^2}{\sqrt{x^2+1}}

Tambua $u$ na $v$ .
Hapa, $u = x^2$ na $v = \sqrt{x^2+1} = (x^2+1)^{1/2}$ .
Tafuta $\frac{du}{dx}$ na $\frac{dv}{dx}$ .
Kwa $u = x^2$ :
$\frac{du}{dx} = 2x$
Kwa $v = (x^2+1)^{1/2}$ (kwa kutumia kanuni ya mnyororo):
$\frac{dv}{dx} = \frac{1}{2}(x^2+1)^{-1/2} \cdot (2x)$ $\frac{dv}{dx} = x(x^2+1)^{-1/2} = \frac{x}{\sqrt{x^2+1}}$
Tumia kanuni ya quotient.
$\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$
$\frac{dy}{dx} = \frac{\sqrt{x^2+1}(2x) - x^2\left(\frac{x}{\sqrt{x^2+1}}\right)}{(\sqrt{x^2+1})^2}$
Rahisisha denominator.
$(\sqrt{x^2+1})^2 = x^2+1$
Rahisisha nambari (numerator).
Nambari: $2x\sqrt{x^2+1} - \frac{x^3}{\sqrt{x^2+1}}$
Pata denominator ya kawaida $\sqrt{x^2+1}$ :
$\text{Nambari} = \frac{2x(x^2+1) - x^3}{\sqrt{x^2+1}}$ $\text{Nambari} = \frac{2x^3+2x - x^3}{\sqrt{x^2+1}}$ $\text{Nambari} = \frac{x^3+2x}{\sqrt{x^2+1}} = \frac{x(x^2+2)}{\sqrt{x^2+1}}$
Unganisha nambari na denominator.
$\frac{dy}{dx} = \frac{\frac{x(x^2+2)}{\sqrt{x^2+1}}}{x^2+1}$ $\frac{dy}{dx} = \frac{x(x^2+2)}{\sqrt{x^2+1}(x^2+1)}$ $\frac{dy}{dx} = \frac{x(x^2+2)}{(x^2+1)^{3/2}}$
Jibu la mwisho ni:
$\boxed{\frac{dy}{dx} = \frac{x(x^2+2)}{(x^2+1)^{3/2}}}$

Find dy/dx for the following: y=x2+1, (b) y=(x2−1)/(2x+1), (c) y=sqrt(x)−1/sqrt(x), (d) y=3x/2x+1, (e) y=x/(x+4)3, (f) y=sqrt(x2−1), (g) y=x2/sqrt(2x+1), (h) y=-1/(1+sqrt(x))2, (i) y=1/(x2+2x-4)

Step-by-step solution

Tambua $u$ na $v$ .

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

Tumia kanuni ya quotient.

Rahisisha denominator.

Rahisisha nambari (numerator).

Unganisha nambari na denominator.

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