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Calculus Level 1

d d x sin ⁡ x = cos ⁡ x d d x cos ⁡ x = sin ⁡ x d d x sinh ⁡ x = cosh ⁡ x d d x cosh ⁡ x = sinh ⁡ x \large{\begin{array} {l c l } \frac{d}{dx} \sin x &= &\cos x \\ \frac{d}{dx} \cos x &= &\sin x \\ \frac{d}{dx} \sinh x &= &\cosh x \\ \frac{d}{dx} \cosh x &= &\sinh x \end{array}}

One of the above equations is wrong, which one?

d d x sin ⁡ x = cos ⁡ x \frac{d}{dx} \sin x = \cos x d d x cosh ⁡ x = sinh ⁡ x \frac{d}{dx} \cosh x = \sinh x d d x sinh ⁡ x = cosh ⁡ x \frac{d}{dx} \sinh x = \cosh x d d x cos ⁡ x = sin ⁡ x \frac{d}{dx} \cos x = \sin x

Pi Han Goh by Pi Han Goh , 30, Malaysia

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1 solution

Ashish Menon
Aug 11, 2016

d d x cos ⁡ x = − sin ⁡ x \dfrac{d}{dx} \cos x = -\sin x , so this \color{#3D99F6}{\boxed{\text{this}}} is wrong.

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Got a proof?

Pi Han Goh - 4 years, 10 months ago

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We can use the definition of derivative to prove this.

d d x cos ⁡ ( x ) = lim ⁡ h → 0 cos ⁡ ( x + h ) − cos ⁡ ( h ) h = lim ⁡ h → 0 cos ⁡ ( x ) cos ⁡ ( h ) − 1 2 − sin ⁡ ( x ) sin ⁡ ( h ) h = ( − cos ⁡ ( x ) ) lim ⁡ h → 0 sin ⁡ 2 ( h / 2 ) h / 2 − sin ⁡ ( x ) sin ⁡ ( h ) h = ( − cos ⁡ ( x ) ) ( lim ⁡ u = ( h / 2 ) → 0 sin ⁡ ( u ) u ) ( lim ⁡ u = ( h / 2 ) → 0 sin ⁡ ( u ) ) − sin ⁡ ( x ) lim ⁡ h → 0 sin ⁡ ( h ) h = ( − cos ⁡ ( x ) ) ⋅ 1 ⋅ 0 − sin ⁡ ( x ) ⋅ 1 = ( − sin ⁡ ( x ) ) \begin{aligned}\frac{\mathrm d}{\mathrm dx}\cos(x)&=\lim_{h\to 0}\frac{\cos(x+h)-\cos(h)}{h}\\&=\lim_{h\to 0}\cos(x)\frac{\cos(h)-1}{2}-\sin(x)\frac{\sin(h)}{h}\\&=(-\cos(x))\lim_{h\to 0}\frac{\sin^2(h/2)}{h/2}-\sin(x)\frac{\sin(h)}{h}\\&=(-\cos(x))\left(\lim_{u=(h/2)\to 0}\frac{\sin(u)}{u}\right)\left(\lim_{u=(h/2)\to 0}\sin(u)\right)-\sin(x)\lim_{h\to 0}\frac{\sin(h)}{h}\\&=(-\cos(x))\cdot 1\cdot 0-\sin(x)\cdot 1=(-\sin(x))\end{aligned}

Here, the result lim ⁡ x → 0 sin ⁡ ( x ) x = 1 \lim\limits_{x\to 0}\frac{\sin(x)}{x}=1 that we used has a very well-known elegant geometric proof using the squeeze theorem. The trigonometric formulas used also have well-known trivial proofs.

Prasun Biswas - 4 years, 10 months ago

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ONE THE FULL!!!!!

Pi Han Goh - 4 years, 10 months ago

Hmm, yeah, thats pretty simple.

d d x cos ⁡ x = d d x sin ⁡ ( π 2 − x ) = cos ⁡ ( π 2 − x ) ( 0 − 1 ) = − cos ⁡ ( π 2 − x ) = − sin ⁡ x \begin{aligned} \dfrac{d}{dx} \cos x & = \dfrac{d}{dx} \sin\left(\dfrac{\pi}{2} - x\right)\\ \\ & = \cos\left(\dfrac{\pi}{2} - x\right)\left(0-1\right)\\ \\ & = -\cos\left(\dfrac{\pi}{2} - x\right)\\ \\ & = \color{#3D99F6}{\boxed{-\sin x}} \end{aligned}

Ashish Menon - 4 years, 10 months ago

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Your working shows that we need to know that d d x sin ⁡ x = cos ⁡ x \frac d{dx} \sin x = \cos x .

That begs the question: If you want to prove that d d x cos ⁡ x = − sin ⁡ x \frac d{dx} \cos x= - \sin x , then you need to prove that d d x sin ⁡ x = cos ⁡ x \frac d{dx} \sin x = \cos x as well. Now, can you prove that d d x sin ⁡ x = cos ⁡ x \frac d{dx} \sin x = \cos x ?

Pi Han Goh - 4 years, 10 months ago

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