question 3\n(a) determine $lim_{x \to 0}\frac{cos x - 1}{2sin x}$\n(b) differentiate the following with…

question 3\n(a) determine $lim_{x \to 0}\frac{cos x - 1}{2sin x}$\n(b) differentiate the following with respect to x:\n(i) $y=ln(2x^{2}+1)$.\n(ii) $y = \frac{e^{x}+1}{e^{x}-1}$\n(iii) $y = x^{4}\tan3x$\n(c) the period $t$ of a simple pendulum of length $l$ is given by $t = 2pisqrt{\frac{l}{g}}$, where $g$ is a constant called the acceleration due to gravity. find $\frac{dt}{dl}$.\n(d) find the turning point of the function $y = x^{4}-8x^{2}+16$ and determine their nature.

question 3\n(a) determine $lim_{x \to 0}\frac{cos x - 1}{2sin x}$\n(b) differentiate the following with respect to x:\n(i) $y=ln(2x^{2}+1)$.\n(ii) $y = \frac{e^{x}+1}{e^{x}-1}$\n(iii) $y = x^{4}\tan3x$\n(c) the period $t$ of a simple pendulum of length $l$ is given by $t = 2pisqrt{\frac{l}{g}}$, where $g$ is a constant called the acceleration due to gravity. find $\frac{dt}{dl}$.\n(d) find the turning point of the function $y = x^{4}-8x^{2}+16$ and determine their nature.

Answer

Explanation:

Step1: Solve part (a)

Use L - H rule as $\lim_{x\rightarrow0}\frac{\cos x - 1}{2\sin x}$ is in $\frac{0}{0}$ form. Differentiate numerator and denominator. $\lim_{x\rightarrow0}\frac{\cos x - 1}{2\sin x}=\lim_{x\rightarrow0}\frac{-\sin x}{2\cos x}=0$

Step2: Solve part (b)(i)

Use chain - rule. If $y = \ln(u)$ and $u = 2x^{2}+1$, then $\frac{dy}{du}=\frac{1}{u}$ and $\frac{du}{dx}=4x$. So $\frac{dy}{dx}=\frac{4x}{2x^{2}+1}$.

Step3: Solve part (b)(ii)

Use quotient - rule $\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}$. Here $u = e^{x}+1$, $u'=e^{x}$, $v = e^{x}-1$, $v'=e^{x}$. $\frac{dy}{dx}=\frac{e^{x}(e^{x}-1)-e^{x}(e^{x}+1)}{(e^{x}-1)^{2}}=\frac{e^{2x}-e^{x}-e^{2x}-e^{x}}{(e^{x}-1)^{2}}=-\frac{2e^{x}}{(e^{x}-1)^{2}}$

Step4: Solve part (b)(iii)

Use product - rule $(uv)' = u'v+uv'$. Here $u = x^{4}$, $u' = 4x^{3}$, $v=\tan3x$, $v' = 3\sec^{2}3x$. $\frac{dy}{dx}=4x^{3}\tan3x + 3x^{4}\sec^{2}3x$

Step5: Solve part (c)

Given $T = 2\pi\sqrt{\frac{l}{g}}=\frac{2\pi}{\sqrt{g}}l^{\frac{1}{2}}$. Differentiate with respect to $l$. $\frac{dT}{dl}=\frac{\pi}{\sqrt{gl}}$

Step6: Solve part (d)

First, find the derivative $y'=4x^{3}-16x = 4x(x^{2}-4)=4x(x - 2)(x + 2)$. Set $y' = 0$, then $x=0,2,-2$. Find the second - derivative $y'' = 12x^{2}-16$. When $x = 0$, $y''=-16<0$, so $(0,16)$ is a local maximum. When $x = 2$, $y''=12\times4 - 16 = 32>0$, so $(2,0)$ is a local minimum. When $x=-2$, $y''=12\times4 - 16 = 32>0$, so $(-2,0)$ is a local minimum.

Answer:

(a) $0$ (b)(i) $\frac{4x}{2x^{2}+1}$ (b)(ii) $-\frac{2e^{x}}{(e^{x}-1)^{2}}$ (b)(iii) $4x^{3}\tan3x + 3x^{4}\sec^{2}3x$ (c) $\frac{\pi}{\sqrt{gl}}$ (d) Turning points are $(0,16)$ (local maximum), $(2,0)$ (local minimum), $(-2,0)$ (local minimum)