We read your photo as:

Find the derivatives of the following with first principle: 1) f(x) = x2−x 2) f(x) = sqrt(x+1) 3) g(x) = 1/x2 4) h(x) = 1/sqrt(x) 5) y = 1/(2x-1)

Mathematics

ScanSolve AI Answer

Step-by-step solution

24 steps

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

1) f(x)=x2−xf(x) = x^2 - x

  1. Write the definition of the derivative.

    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+h)f(x+h) and f(x)f(x) into the formula.

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

  3. Expand and simplify the numerator.

    f′(x)=lim⁡h→0(x2+2xh+h2−x−h)−(x2−x)hf'(x) = \lim_{h \to 0} \frac{(x^2 + 2xh + h^2 - x - h) - (x^2 - x)}{h} f′(x)=lim⁡h→0x2+2xh+h2−x−h−x2+xhf'(x) = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x - h - x^2 + x}{h} f′(x)=lim⁡h→02xh+h2−hhf'(x) = \lim_{h \to 0} \frac{2xh + h^2 - h}{h}

  4. Factor out hh from the numerator and cancel it with the denominator.

    f′(x)=lim⁡h→0h(2x+h−1)hf'(x) = \lim_{h \to 0} \frac{h(2x + h - 1)}{h} f′(x)=lim⁡h→0(2x+h−1)f'(x) = \lim_{h \to 0} (2x + h - 1)

  5. Evaluate the limit by substituting h=0h=0.

    f′(x)=2x+0−1=2x−1f'(x) = 2x + 0 - 1 = 2x - 1

    The derivative is 2x−1\boxed{2x-1}.

    2) f(x)=x+1f(x) = \sqrt{x+1}

  6. Write the definition of the derivative.

    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+h)f(x+h) and f(x)f(x) into the formula.

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

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

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

  9. Simplify the numerator using (a−b)(a+b)=a2−b2(a-b)(a+b)=a^2-b^2.

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

  10. Cancel out hh and evaluate the limit by substituting h=0h=0.

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

    The derivative is 12x+1\boxed{\frac{1}{2\sqrt{x+1}}}.

    3) g(x)=1x2g(x) = \frac{1}{x^2}

  11. Write the definition of the derivative.

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

  12. Substitute g(x+h)g(x+h) and g(x)g(x) into the formula.

    g′(x)=lim⁡h→01(x+h)2−1x2hg'(x) = \lim_{h \to 0} \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h}

  13. Combine the fractions in the numerator.

    g′(x)=lim⁡h→0x2−(x+h)2x2(x+h)2hg'(x) = \lim_{h \to 0} \frac{\frac{x^2 - (x+h)^2}{x^2(x+h)^2}}{h} g′(x)=lim⁡h→0x2−(x2+2xh+h2)hx2(x+h)2g'(x) = \lim_{h \to 0} \frac{x^2 - (x^2 + 2xh + h^2)}{h x^2(x+h)^2} g′(x)=lim⁡h→0−2xh−h2hx2(x+h)2g'(x) = \lim_{h \to 0} \frac{-2xh - h^2}{h x^2(x+h)^2}

  14. Factor out hh from the numerator and cancel it with the denominator.

    g′(x)=lim⁡h→0h(−2x−h)hx2(x+h)2g'(x) = \lim_{h \to 0} \frac{h(-2x - h)}{h x^2(x+h)^2} g′(x)=lim⁡h→0−2x−hx2(x+h)2g'(x) = \lim_{h \to 0} \frac{-2x - h}{x^2(x+h)^2}

  15. Evaluate the limit by substituting h=0h=0.

    g′(x)=−2x−0x2(x+0)2=−2xx2⋅x2=−2xx4g'(x) = \frac{-2x - 0}{x^2(x+0)^2} = \frac{-2x}{x^2 \cdot x^2} = \frac{-2x}{x^4} g′(x)=−2x3g'(x) = -\frac{2}{x^3}

    The derivative is −2x3\boxed{-\frac{2}{x^3}}.

    4) h(x)=1xh(x) = \frac{1}{\sqrt{x}}

  16. Write the definition of the derivative.

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

  17. Substitute h(x+h)h(x+h) and h(x)h(x) into the formula.

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

  18. Combine the fractions in the numerator.

    h′(x)=lim⁡h→0x−x+hxx+hh=lim⁡h→0x−x+hhxx+hh'(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}}

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

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

  20. Cancel out hh and evaluate the limit by substituting h=0h=0.

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

    The derivative is −12xx\boxed{-\frac{1}{2x\sqrt{x}}}.

    5) y=12x−1y = \frac{1}{2x-1}

  21. Write the definition of the derivative.

    dydx=lim⁡h→0f(x+h)−f(x)h\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

  22. Substitute f(x+h)f(x+h) and f(x)f(x) into the formula.

    dydx=lim⁡h→012(x+h)−1−12x−1h\frac{dy}{dx} = \lim_{h \to 0} \frac{\frac{1}{2(x+h)-1} - \frac{1}{2x-1}}{h} dydx=lim⁡h→012x+2h−1−12x−1h\frac{dy}{dx} = \lim_{h \to 0} \frac{\frac{1}{2x+2h-1} - \frac{1}{2x-1}}{h}

  23. Combine the fractions in the numerator.

    dydx=lim⁡h→0(2x−1)−(2x+2h−1)(2x+2h−1)(2x−1)h\frac{dy}{dx} = \lim_{h \to 0} \frac{\frac{(2x-1) - (2x+2h-1)}{(2x+2h-1)(2x-1)}}{h} dydx=lim⁡h→02x−1−2x−2h+1h(2x+2h−1)(2x−1)\frac{dy}{dx} = \lim_{h \to 0} \frac{2x-1-2x-2h+1}{h(2x+2h-1)(2x-1)} dydx=lim⁡h→0−2hh(2x+2h−1)(2x−1)\frac{dy}{dx} = \lim_{h \to 0} \frac{-2h}{h(2x+2h-1)(2x-1)}

  24. Cancel out hh and evaluate the limit by substituting h=0h=0.

    dydx=lim⁡h→0−2(2x+2h−1)(2x−1)\frac{dy}{dx} = \lim_{h \to 0} \frac{-2}{(2x+2h-1)(2x-1)} dydx=−2(2x+2(0)−1)(2x−1)=−2(2x−1)(2x−1)\frac{dy}{dx} = \frac{-2}{(2x+2(0)-1)(2x-1)} = \frac{-2}{(2x-1)(2x-1)} dydx=−2(2x−1)2\frac{dy}{dx} = -\frac{2}{(2x-1)^2}

    The derivative is −2(2x−1)2\boxed{-\frac{2}{(2x-1)^2}}.

Was this helpful?

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Mathematics Questions

QAssignment on Differentiation, find the derivation of the following with x using first principle: 1…QFind the derivatives of the following functions from first principles or by using differentiation…QFind the derivatives of the following functions using first principles: f(x) = x3 and f(x)…QFind the derivative of the following functions from first principles: i) y = x2 ii) y = 3x + 2 iii)…QI) Define derivatives of a function from first principle. ii) Find the derivatives of f(x) = 3x2…
Browse by topic:Calculus
See all Mathematics questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp