\(f(x) = (5x-8)^{12}\)

\(f'(x) = 12(5x-8)^{11} \dfrac{d}{dx} (5x-8)\)

\(= 12(5x - 8)^{11}.5\)

\(= 60(5x-8)^{11}\)

\(g(x) = \sin (x^2 + 7x-9)\)

\(g'(x) = \cos (x^2 + 7x - 9) (2x+7)\)

\(= (2x+7)\cos (x^2 + 7x - 9)\)

\(h(x) = \sqrt{x^2 + 4}\)

\(h'(x) = \dfrac{1}{2}(x^2 + 4)^{-1/2}(2s)\)

\(= \dfrac{x}{\sqrt{x^2 + 4}}\)

\(f(x) = x^3 \tan (x^2 + 3x)\)

Using the product rule,

\(f'(x) = x^3 \dfrac{d}{dx}(\tan (x^2 + 3x)) + \tan(x^2 + 3x)(3x^2)\)

\(= x^3 \sec^2 (x^2 + 3x)(2x+3) + 3x^2 \tan(x^2 + 3x)\)

\(= (2x^4 + 3x^3)\sec^2 (x^2 + 3x) + 3x^2 + \tan(x^2 + 3x)\)

\(h(x) = \dfrac{2}{\sqrt[3]{3x+4}}\)

\(h(x) = 2(3x+4)^{-1/3}\)

\(h'(x) = 2 * \dfrac{-1}{3} * (3x + 4)^{-1/3 - 1}.*h3\)

\(= -2(3x+4)^{-4/3}\)

\(= \dfrac{-2}{(3x+4)^{4/3}}\)

\(= \dfrac{-2}{\sqrt[3]{(3x+4)^4}}\)

\(g(x) = \sec^3 (x^2 - 5x)\)

\(g(x) = (\sec (x^2 - 5x))^3\)

\(g'(x) = 3\sec^2 (x^2 - 5x) \dfrac{d}{dx} (\sec (x^2 - 5x))\)

\(= 3\sec^2 (x^2 - 5x) \sec (x^2 - 5x) \tan (x^2 - 5x) . 2x\)

\(= 6x \sec^3 (x^2 - 5x) \tan (x^2 - 5x)\)

\(f(x) = \cos^2 (3x)\)

\(f'(x) = 2\cos (3x) - \sin (3x) 3\)

\(= -6 \sin 3x \cos 3x\)

\(g(x) = \sqrt{\cos^2 (3x) + 1}\)

\(g'(x) = \dfrac{1}{2} (\cos^2 (3x) + 1)^{-1/2} (-6\sin 3x \cos 3x)\) (From solution 7)

\(= \dfrac{-3 \sin 3x \cos 3x}{\sqrt{\cos^2 (3x) + 1}}\)

\(h(x) = x\sqrt[4]{2x + 3}\)

\(h(x) = x(2x+3)^{1/4}\)

Using the product rule,

\(h'(x) = x \dfrac{d}{dx}(2x+3)^{1/4} + (2x+3)^{1/4}\)

\(= \dfrac{x}{4}(2x+3)^{-3/4}2 + (2x+3)^{1/4}\)

\(= \dfrac{x}{2\sqrt[4]{(2x+3)^3}} + \sqrt[4]{(2x+3)}\)

\(f(x) = x^2\sqrt[3]{3x-5}\)

\(f(x) = x^2(3x-5)^{1/3}\)

Using the product rule,

\(f'(x) = x^2\dfrac{d}{dx}(3x-5)^{1/3} + 2x(3x-5)^{1/3}\)

\(= x^2\dfrac{1}{3}(3x-5)^{-2/3} + 2x(3x-5)^{1/3}\)

\(= \dfrac{x^2}{3\sqrt[3]{(3x-5)^2}} + 2x\sqrt[3]{3x-5}\)

\(g(x) = \sin (\cos (\tan (3x)))\)

\(g'(x) = \cos (\cos (\tan (3x))) \dfrac{d}{dx} (\cos (\tan 3x))\)

\(= \cos (\cos (\tan 3x)) * - \sin (\tan 3x) \dfrac{d}{dx} (\tan 3x)\)

\(= \cos (\cos (\tan 3x)) * -\sin (\tan 3x) \sec^2 3x 3\)

\(= -3 \sec^2 3x \sin(\tan 3x) \cos (\cos (\tan 3x))\)

\(h(x) = \sqrt{5 + 4\sqrt{3 + 2\sqrt{1+x}}}\)

\(h(x) = (5 + 4\sqrt{3 + 2\sqrt{1+x}})^{1/2}\)

\(h'(x) = \dfrac{1}{2}(5 + 4\sqrt{3 + 2\sqrt{1+x}})^{-1/2} \dfrac{d}{dx}(4\sqrt{3 + 2\sqrt{1+x}})\)

\(= \dfrac{1}{2}(5+ 4\sqrt{3 + 2\sqrt{1+x}})^{-1/2} . 4(3+2\sqrt{1+2})^{-1/2} \dfrac{d}{dx} (2\sqrt{1+x})\)

\(= 2(5+ 4\sqrt{3 + 2\sqrt{1+x}})^{-1/2} 4\sqrt{3 + 2\sqrt{1+x}} 2(1+x)^{-1/2}\)

\(= \dfrac{4}{\sqrt{(5 + 4 \sqrt{3+2\sqrt{1+x}})}(\sqrt{3 + 2\sqrt{1+x}})\sqrt{1+x}}\)

\(f(x) = (\dfrac{2x+5}{2-5x})^7\)

\(f(x) = (2x+5)^2(2-5x)^{-7}\)

Using the product rule,

\(f'(x) = (2x+5)^7\dfrac{d}{dx}(2-5x)^{-7} + (2-5x)^{-7} \dfrac{d}{dx} (2x+5)^7\)

\(= (2x+5)^7.-7(2-5x)^{-8}(-5) + (2-5x)^{-7}.7(2x+5)^6.2\)

\(= \dfrac{35(2x+5)^7}{(2-5x)^8} + \dfrac{14(2x+5)^6}{(2-5x)^7}\)

\(g(x) = \dfrac{\cos 5x}{\sin 3x}\)

\(g(x) = \cos 5x \csc 3x\)

Using the product rule,

\(g'(x) = \cos 5x \dfrac{d}{dx} \csc (3x) + \csc (3x) \dfrac{d}{dx} (\cos 5x)\)

\(= \cos 5x - \csc 3x \cot 3x . 3 + \csc 3x (-\sin 5x) . 5\)

\(= -3\cos 5x \csc 3x - \cot 3x - 5\sin 5x \csc 3x\)

\(f(x) = \sqrt{\tan x}\)

\(f'(x) = \dfrac{1}{2} (\tan x)^{-1/2} \sec^2 x\)

\(= \dfrac{\sec^2 x}{2 \sqrt{\tan x}}\)

\(g(x) = \sqrt{\dfrac{x^2 - 1}{x^2 + 1}}\)

\(g(x) = (x^2 - 1)^{1/2}(x^2 + 1)^{-1/2}\)

Using the product rule,

\(g'(x) = (x^2-1)^{1/2}\dfrac{d}{dx}(x^2+1)^{-1/2}+(x^2+1)^{-1/2}\dfrac{d}{dx}(x^2-1)^{1/2}\)

\(= \dfrac{-1}{2}(x^2-1)^{1/2}(x^2+1)^{-3/2}(2x) + \dfrac{1}{2}(x^2+1)^{-1/2}(x^2-1)^{-1/2}(2x)\)

\(= \dfrac{-x\sqrt{x^2-1}}{(x^2+1)^{3/2}} + \dfrac{x}{\sqrt{(x^2+1)(x^2-2)}}\)

\(g(x) = f(x^2-3x+1)\)

\(f'(x)=3\)

\(f''(1) = -1\)

Let \(h=x^2-3x+1\)

\(g(x) = f \circ h\)

Using chain rule,

\(g'(x) = f'(h(x))h'(x)\)

\(h'(x) = 2x- 3\)

\(g'(3) = f'(f(3))h'(3)\)

\(h(3) = 3^2 - 3^2 + 1 = 1\)

\(h'(3) = 3.2 - 3 = 3\)

\(g'(3) = f'(1)h'(3)\)

\(= 3.3 = 9\)

We know that,

\(g'(x) = f'(h(x))h'(x)\)

Using the product rule,

\(g''(x) = f'(h(x))h''(x) + h'(x)\dfrac{d}{dx}(f'(h(x)))\)

\(= f'(h(x))h''(x) + h'(x)f''(h(x))h'(x)\)

\(= f'(h(x))h''(x) + (h'(x))^2f''(h(x))\)

\(g''(3) = f'(h(3))h''(3) + (h'(3))^2f''(h(3))\)

\(h(3) = 1\)

\(h'(3) = 3\)

\(h''(x) = 2\)

\(g''(3) = f'(1).2 + 3^2 f''(1)\)

\(= 2f'(1)+9f''(1)\)

\(=2.3 + 9(-1)\)

\(= 6 - 9 = -3\)

\(y = 5x - 3\) is tangent to \(y=f(x)\) at \((3,12)\)

\(g(x) = \dfrac{f(x^2-1)}{x}\)

We need to find tangent for \(y=g(x)\) at \((2,g(x))\)

\(g(2) = \dfrac{f(4-1)}{3} = \dfrac{f(3)}{2}\)

We know that \(f(3) = 12\)

\(g(2) = \dfrac{12}{2} = 6\)

So we need to find tangent to \(y=g(x)\) at \((2,6\))

\(g(x) = \dfrac{f(x^2-1)}{2}\)

\(g'(x) = \dfrac{1}{2} \dfrac{d}{dx}f(x^2-1)\)

\(= \dfrac{1}{2}f'(x^2-1)(2x)\)

\(= xf'(x^2-1)\)

\(g'(2) = 2f'(3) = 2.12 = 24\)

So we have the following data about the tangent:

It's point: (2.6)

It's slope = 24

\(y = mx + c\)

\(6 = 24.2 + c\)

\(c = 6 - 48 = -42\)

\(y = 24x - 42\)

\(f(x) = (3x-2)^7\)

\(f'(x) = 7(3x-2)^6(3)\)

\(= 21(3x-2)^6\)

\(f''(x) = 126(3x-2)^5\)

\(= 378(3x-2)^5\)

\(f'''(x) = 378.5(3x-2)^4\)

\(= 1890(3x-2)^4\)

\(g(x) = \sec(3x + 1)\)

\(g'(x) = \sec(3x + 1) \tan(3x+1).e\)

\(g''(x) = 3\sec(3x+1)\dfrac{d}{dx}\tan(3x+1) + 3\tan(3x+1)\dfrac{d}{dx}\sec(3x+1)\)

\(= 3\sec(3x+1)\sec^2(3x+1).3 + 3\tan(3x+1)\sec(3x+1)\tan(3x+1).3\)

\(= 9\sec^3(3x + 1) + 9\tan^2(3x+1)\sec(3x + 1)\)

\(g'''(x) = 27\sec^2(3x+1).3 + 9\tan^2(3x+1)\dfrac{d}{dx}\sec(3x+1) + 9\sec(3x+1)\dfrac{d}{dx}\tan^2(3x+1)\)

\(= 27\sec^2(3x+1) + 9\tan^2(3x+1)\sec(3x+1)\tan(3x+1).3 + 9\sec(3x+1).2\tan(3x+1).3\sec^2(3x+1)\)

\(= 27\sec^2(3x+1) + 27\tan^3(3x+1)\sec(3x+1) + 54\sec^3(3x+1)\tan(3x+1)\)

\(h(x) = \sin(2x)\)

\(h'(x) = \cos(2x).2 = 2\cos (2x)\)

\(h''(x) = -4\sin(2x)\)

\(h'''(x) = -8\cos(2x)\)

\(h''''(x) = 16\sin(2x)\)

So looking at the pattern, when we do differentiate 100th time,

\(h^{(100)}x = 2^{100}\sin(2x)\)

\(y = \sin(\pi t^2)\)

Velocity = \(\dfrac{dy}{dt} = \cos (\pi t^2).2t\pi\)

\(= 2t\pi\cos(\pi t^2)\)

Acceleration = \(\dfrac{d^2y}{dt^2} = 2t\pi \dfrac{d}{dt} (\cos \pi t^2) + (\cos \pi t^2 \dfrac{d}{dt} (2t\pi))\)

\(= -2\pi t\sin(\pi t^2)2\pi t + \pi t^2 (2\pi)\)

\(= -4\pi^2t^2 \sin(\pi t^2) + 2\pi^2 t^2\)

\(= 2\pi^2 t^2 - 4\pi^2t^2\sin(\pi t^2)\)

At \(t=1\),

Velocity = \(2\pi(cos \pi) = -2\pi\)

Acceleration = \(2\pi^2 - 4\pi^2\sin(\pi)\)

\(= 2\pi^2 - 0 = 2\pi^2\)

\(\forall x \sin(2x) = 2\sin x \cos x\)

\(2 \cos 2x = 2 \sin \dfrac{d}{dx}\cos x + 2 \cos x \dfrac{d}{dx} \sin x\)

\(2\cos 2x = -2\sin x \sin x + 2\cos x \cos x\)

\(2\cos 2x = -2\sin^2 x + 2\cos^2 x\)

\(\cos 2x = \cos^2x - \sin^2 x\)

\(f(x) = x^{m/n}\)

\(m\) and \(n\) are positive integers.

\(m > n \geq 3\)

\(n\) is odd.

\(m/n\) is reduced to lowest terms

\(y = x^{m/n}\)

\(y^n = x^m\)

Differentiating above w.r.t x,

\(\dfrac{d}{dx}y^n = mx^{m-1}\)

\(ny^{n-1}\dfrac{dy}{dx} = mx^{m-1}\)

\(\dfrac{dy}{dx} = \dfrac{m(x^{m-1})}{ny^{n-1}}\)

\(= \dfrac{mx^{m-1}}{n(x^{m/n})^{n-1}}\)

\(= \dfrac{m}{n}x^{m-1-\dfrac{m(n-1)}{n}}\)

\(= \dfrac{m}{n}x^{\dfrac{m}{n}-1}\)

Now since \(m/n > 1\), we know that \(m/n - 1 > 0\).

So when \(x=0\), the derivative is defined.

\(f'(0) = \dfrac{m}{n} 0^{m/n} = 0\)