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Assignment. Use first principles to find the derivative of: (i) f(x) = √x (ii) f(x) = 1/√x (iii) f(x) = Sinx (iv) f(x) = Cosx

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4 parts
i

f′(x)=12xf'(x) = \frac{1}{2\sqrt{x}}

ii

f′(x)=−12xxf'(x) = -\frac{1}{2x\sqrt{x}}

iii

f′(x)=cos⁡xf'(x) = \cos x

iv

f′(x)=−sin⁡xf'(x) = -\sin x

Step-by-step solution

13 steps

Here are the solutions to the assignment, finding the derivative of each function using first principles. The formula for the derivative from first principles is:

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

Part i· 3 steps

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

  1. Write f(x+h)f(x+h) and substitute into the formula.

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

  2. Multiply the numerator and denominator by the conjugate of the numerator.

    f′(x)=lim⁡h→0x+h−xh×x+h+xx+h+xf'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \times \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}} f′(x)=lim⁡h→0(x+h)−xh(x+h+x)f'(x) = \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})}

  3. Simplify the expression and evaluate the limit.

    f′(x)=lim⁡h→0hh(x+h+x)f'(x) = \lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})} f′(x)=lim⁡h→01x+h+xf'(x) = \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}} f′(x)=1x+0+x=12xf'(x) = \frac{1}{\sqrt{x+0} + \sqrt{x}} = \frac{1}{2\sqrt{x}}

    The derivative is:

    f′(x)=12x\boxed{f'(x) = \frac{1}{2\sqrt{x}}}

Part ii

f′(x)=−12xxf'(x) = -\frac{1}{2x\sqrt{x}}

f(x)=1xf(x) = \frac{1}{\sqrt{x}}

  1. Write f(x+h)f(x+h) and substitute into the formula.

    f′(x)=lim⁡h→01x+h−1xhf'(x) = \lim_{h \to 0} \frac{\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}}{h}

  2. Combine the fractions in the numerator.

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

  3. Multiply the numerator and denominator by the conjugate of the numerator.

    f′(x)=lim⁡h→0x−x+hhxx+h×x+x+hx+x+hf'(x) = \lim_{h \to 0} \frac{\sqrt{x} - \sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}} \times \frac{\sqrt{x} + \sqrt{x+h}}{\sqrt{x} + \sqrt{x+h}} f′(x)=lim⁡h→0x−(x+h)hxx+h(x+x+h)f'(x) = \lim_{h \to 0} \frac{x - (x+h)}{h\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})} f′(x)=lim⁡h→0−hhxx+h(x+x+h)f'(x) = \lim_{h \to 0} \frac{-h}{h\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})}

  4. Simplify the expression and evaluate the limit.

    f′(x)=lim⁡h→0−1xx+h(x+x+h)f'(x) = \lim_{h \to 0} \frac{-1}{\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})} f′(x)=−1xx(x+x)=−1x(2x)=−12x3/2f'(x) = \frac{-1}{\sqrt{x}\sqrt{x}(\sqrt{x} + \sqrt{x})} = \frac{-1}{x(2\sqrt{x})} = -\frac{1}{2x^{3/2}}

    The derivative is:

    f′(x)=−12xx\boxed{f'(x) = -\frac{1}{2x\sqrt{x}}}

Part iii

f′(x)=cos⁡xf'(x) = \cos x

f(x)=sin⁡xf(x) = \sin x

  1. Write f(x+h)f(x+h) and substitute into the formula.

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

  2. Use the trigonometric identity sin⁡A−sin⁡B=2cos⁡(A+B2)sin⁡(A−B2)\sin A - \sin B = 2 \cos\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→02cos⁡(x+h+x2)sin⁡(x+h−x2)hf'(x) = \lim_{h \to 0} \frac{2 \cos\left(\frac{x+h+x}{2}\right) \sin\left(\frac{x+h-x}{2}\right)}{h} f′(x)=lim⁡h→02cos⁡(x+h2)sin⁡(h2)hf'(x) = \lim_{h \to 0} \frac{2 \cos\left(x + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)}{h}

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

    f′(x)=lim⁡h→0(cos⁡(x+h2)×sin⁡(h2)h2)f'(x) = \lim_{h \to 0} \left( \cos\left(x + \frac{h}{2}\right) \times \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right) f′(x)=cos⁡(x+0)×1f'(x) = \cos(x+0) \times 1

    The derivative is:

    f′(x)=cos⁡x\boxed{f'(x) = \cos x}

Part iv

f′(x)=−sin⁡xf'(x) = -\sin x

f(x)=cos⁡xf(x) = \cos x

  1. Write f(x+h)f(x+h) and substitute into the formula.

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

  2. Use the trigonometric 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}

  3. Rearrange the terms and 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) \times \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right) f′(x)=−sin⁡(x+0)×1f'(x) = -\sin(x+0) \times 1

    The derivative is:

    f′(x)=−sin⁡x\boxed{f'(x) = -\sin x}
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