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Write the definition of the derivative from first principles.

Mathematics

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Answer

f′(x)=−sin⁡xf'(x) = \mathbf{-\sin x}

f′(x)=sec⁡2xf'(x) = \mathbf{\sec^2 x}

Step-by-step solution

11 steps

Here are the derivatives of the given functions using first principles:

1) f(x)=cos⁡xf(x) = \cos x

  1. Write the definition of the derivative from first principles.

    f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

  2. Substitute f(x)=cos⁡xf(x) = \cos x and f(x+h)=cos⁡(x+h)f(x+h) = \cos(x+h) into the formula.

    f′(x)=lim⁡h→0cos⁡(x+h)−cos⁡xhf'(x) = \lim_{h \to 0} \frac{\cos(x+h) - \cos x}{h}

  3. Use the sum-to-product identity cos⁡A−cos⁡B=−2sin⁡(A+B2)sin⁡(A−B2)\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right).

    Here, A=x+hA = x+h and B=xB = x.

    f′(x)=lim⁡h→0−2sin⁡(x+h+x2)sin⁡(x+h−x2)hf'(x) = \lim_{h \to 0} \frac{-2 \sin\left(\frac{x+h+x}{2}\right) \sin\left(\frac{x+h-x}{2}\right)}{h} f′(x)=lim⁡h→0−2sin⁡(x+h2)sin⁡(h2)hf'(x) = \lim_{h \to 0} \frac{-2 \sin\left(x+\frac{h}{2}\right) \sin\left(\frac{h}{2}\right)}{h}

  4. Rearrange the terms to use the standard limit lim⁡θ→0sin⁡θθ=1\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1.

    f′(x)=lim⁡h→0(−sin⁡(x+h2)⋅sin⁡(h2)h2)f'(x) = \lim_{h \to 0} \left( -\sin\left(x+\frac{h}{2}\right) \cdot \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right)

  5. Apply the limit as h→0h \to 0.

    f′(x)=−sin⁡(x+0)⋅1f'(x) = -\sin(x+0) \cdot 1 f′(x)=−sin⁡xf'(x) = \boxed{\mathbf{-\sin x}}

    2) f(x)=tan⁡xf(x) = \tan x

  6. Write the definition of the derivative from first principles.

    f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

  7. Substitute f(x)=tan⁡xf(x) = \tan x and f(x+h)=tan⁡(x+h)f(x+h) = \tan(x+h) into the formula.

    f′(x)=lim⁡h→0tan⁡(x+h)−tan⁡xhf'(x) = \lim_{h \to 0} \frac{\tan(x+h) - \tan x}{h}

  8. Rewrite tan⁡x\tan x as sin⁡xcos⁡x\frac{\sin x}{\cos x} and combine the fractions in the numerator.

    f′(x)=lim⁡h→0sin⁡(x+h)cos⁡(x+h)−sin⁡xcos⁡xhf'(x) = \lim_{h \to 0} \frac{\frac{\sin(x+h)}{\cos(x+h)} - \frac{\sin x}{\cos x}}{h} f′(x)=lim⁡h→0sin⁡(x+h)cos⁡x−cos⁡(x+h)sin⁡xcos⁡(x+h)cos⁡xhf'(x) = \lim_{h \to 0} \frac{\frac{\sin(x+h)\cos x - \cos(x+h)\sin x}{\cos(x+h)\cos x}}{h}

  9. Use the sine subtraction formula sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin(A-B) = \sin A \cos B - \cos A \sin B.

    Here, A=x+hA = x+h and B=xB = x.

    f′(x)=lim⁡h→0sin⁡((x+h)−x)cos⁡(x+h)cos⁡xhf'(x) = \lim_{h \to 0} \frac{\frac{\sin((x+h)-x)}{\cos(x+h)\cos x}}{h} f′(x)=lim⁡h→0sin⁡hhcos⁡(x+h)cos⁡xf'(x) = \lim_{h \to 0} \frac{\sin h}{h \cos(x+h)\cos x}

  10. Separate the limit into known parts.

    f′(x)=lim⁡h→0(sin⁡hh⋅1cos⁡(x+h)cos⁡x)f'(x) = \lim_{h \to 0} \left( \frac{\sin h}{h} \cdot \frac{1}{\cos(x+h)\cos x} \right)

  11. Apply the limit as h→0h \to 0.

    f′(x)=1⋅1cos⁡(x+0)cos⁡xf'(x) = 1 \cdot \frac{1}{\cos(x+0)\cos x} f′(x)=1cos⁡xcos⁡xf'(x) = \frac{1}{\cos x \cos x} f′(x)=1cos⁡2xf'(x) = \frac{1}{\cos^2 x} f′(x)=sec⁡2xf'(x) = \boxed{\mathbf{\sec^2 x}}
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