a) Consider f(x) = 3x − 8 for x ≤ 5, and f(x) = 2k for x 5. Find the value of k if f(x) is continuous at x = 5. (2 marks)
b) Find dy/dx, if y = (Sin x)^(log x), Sin x 0. (2 marks)
c) If y = (Sin⁻¹x)², then show that (1 − x²) d²y/dx² − x dy/dx = 2. (2 marks)

Question

a) Consider f(x) = 3x − 8 for x ≤ 5, and f(x) = 2k for x > 5. Find the value of k if f(x) is continuous at x = 5. (2 marks)
b) Find dy/dx, if y = (Sin x)^(log x), Sin x > 0. (2 marks)
c) If y = (Sin⁻¹x)², then show that (1 − x²) d²y/dx² − x dy/dx = 2. (2 marks)

Accepted Answer

>

> [!TLDR]
> (a) Match the left- and right-hand limits at $x=5$; (b) take logarithms to differentiate $y=(\sin x)^{\log x}$; (c) differentiate $\sqrt{1-x^2}\,y' = 2\sin^{-1}x$ once more.
>
> **a) Continuity at $x=5$**
>
> $f(x) = 3x-8$ for $x\le5$, so $f(5) = 3(5)-8 = 7$, and $\lim_{x	o5^-}f(x)=7$.
> $f(x)=2k$ for $x>5$, so $\lim_{x	o5^+}f(x) = 2k$.
> For continuity at $x=5$: $2k = 7 \implies k = \dfrac72$.
>
> **b) $\dfrac{dy}{dx}$ for $y=(\sin x)^{\log x}$**
>
> Take $\log$ of both sides:
> $$\log y = \log x \cdot \log(\sin x)$$
> Differentiate w.r.t. $x$ (product rule on the right):
> $$\frac1y\frac{dy}{dx} = \frac1x\log(\sin x) + \log x \cdot \frac{\cos x}{\sin x}$$
> $$\frac{dy}{dx} = (\sin x)^{\log x}\left[\frac{\log(\sin x)}{x} + \log x\cdot\cot xight]$$
>
> **c) $y=(\sin^{-1}x)^2$: show $(1-x^2)y'' - xy' = 2$**
>
> $$y' = \frac{2\sin^{-1}x}{\sqrt{1-x^2}} \implies \sqrt{1-x^2}\,y' = 2\sin^{-1}x$$ …

Order	Leibniz	Lagrange	Newton
1st	$\frac{dy}{dx}$	$f'(x)$	$\dot{y}$
2nd	$\frac{d^2y}{dx^2}$	$f''(x)$	$\ddot{y}$
$n$ th	$\frac{d^ny}{dx^n}$	$f^{(n)}(x)$	—

Q.a) Consider f(x) = 3x − 8 for x ≤ 5, and f(x) = 2k for x > 5. Find the value of k if f(x) is continuous at x = 5. (2 marks)

Concept understanding — Successive Differentiation

Successive Differentiation — Repeated Slopes

The definition and notation

Seeing the pattern

Q.a) Consider f(x) = 3x − 8 for x ≤ 5, and f(x) = 2k for x > 5. Find the value of k if f(x) is continuous at x = 5. (2 marks)

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