Differentiate the following with respect to x: 1. y = x^(1/2) + 1/x2 - 3/x3 + 4x^(1/4)

Mathematics
Differentiate the following with respect to x: 1. y = x^(1/2) + 1/x2 - 3/x3 + 4x^(1/4)

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Answer

y′=12x−94x1/4y' = \dfrac{1}{2\sqrt{x}} - \dfrac{9}{4 x^{1/4}}

1.

y=x12−3x34y = x^{\frac{1}{2}} - 3x^{\frac{3}{4}}

Step 1: Power rule ddxxn=nxn−1\frac{d}{dx} x^{n} = n x^{n-1}.

ddx(x12)=12x12−1=12x−12\frac{d}{dx} \left( x^{\frac{1}{2}} \right) = \frac{1}{2} x^{\frac{1}{2}-1} = \frac{1}{2} x^{-\frac{1}{2}}

Step 2:

ddx(−3x34)=−3⋅34x34−1=−94x−14\frac{d}{dx} \left( -3x^{\frac{3}{4}} \right) = -3 \cdot \frac{3}{4} x^{\frac{3}{4}-1} = -\frac{9}{4} x^{-\frac{1}{4}}

Step 3: Sum the derivatives.

y′=12x−12−94x−14y' = \frac{1}{2} x^{-\frac{1}{2}} - \frac{9}{4} x^{-\frac{1}{4}}

y' = \dfrac{1{2\sqrt{x}} - \dfrac{9}{4 x^{1/4}}}

2.

y=x−cos⁡x−2sin⁡xy = x - \cos x - 2\sin x

Step 1: ddxx=1\frac{d}{dx} x = 1, ddx(−cos⁡x)=sin⁡x\frac{d}{dx} (-\cos x) = \sin x, ddx(−2sin⁡x)=−2cos⁡x\frac{d}{dx} (-2\sin x) = -2\cos x.

Step 2: Sum the derivatives.

y′=1+sin⁡x−2cos⁡xy' = 1 + \sin x - 2\cos x

y' = 1 + \sin x - 2\cos x

3.

y=extan⁡xy = e^{x} \tan x

Step 1: Product rule ddx(uv)=u′v+uv′\frac{d}{dx} (uv) = u'v + uv', where u=exu = e^x, v=tan⁡xv = \tan x.

u′=ex,v′=sec⁡2xu' = e^x, \quad v' = \sec^2 x

Step 2: Substitute.

y′=extan⁡x+exsec⁡2xy' = e^x \tan x + e^x \sec^2 x

Step 3: Factor.

y′=ex(tan⁡x+sec⁡2x)y' = e^x (\tan x + \sec^2 x)

y' = e^{x (\tan x + \sec^{2} x)}

4.

y=(ln⁡x)2−4ln⁡xy = (\ln x)^2 - 4 \ln x

Step 1: Chain rule for (ln⁡x)2(\ln x)^2: let u=ln⁡xu = \ln x, dydu=2u\frac{dy}{du} = 2u, dudx=1x\frac{du}{dx} = \frac{1}{x}.

ddx(ln⁡x)2=2ln⁡x⋅1x\frac{d}{dx} (\ln x)^2 = 2\ln x \cdot \frac{1}{x}

Step 2: ddx(−4ln⁡x)=−4⋅1x\frac{d}{dx} (-4 \ln x) = -4 \cdot \frac{1}{x}.

Step 3: Sum.

y′=2ln⁡xx−4x=2ln⁡x−4xy' = \frac{2\ln x}{x} - \frac{4}{x} = \frac{2\ln x - 4}{x}

y' = \dfrac{2\ln x - 4{x}}

5.

y=sin⁡(3x+1)y = \sin(3x + 1)

Step 1: Chain rule ddxsin⁡u=cos⁡u⋅u′\frac{d}{dx} \sin u = \cos u \cdot u', u=3x+1u = 3x + 1, u′=3u' = 3.

y′=cos⁡(3x+1)⋅3=3cos⁡(3x+1)y' = \cos(3x + 1) \cdot 3 = 3 \cos(3x + 1)

y' = 3\cos(3x + 1)

6.

y=ln⁡(sec⁡x)y = \ln(\sec x)

Step 1: Chain rule ddxln⁡u=1uu′\frac{d}{dx} \ln u = \frac{1}{u} u', u=sec⁡xu = \sec x, u′=sec⁡xtan⁡xu' = \sec x \tan x.

y′=1sec⁡x⋅sec⁡xtan⁡xy' = \frac{1}{\sec x} \cdot \sec x \tan x

Step 2: Simplify.

y′=tan⁡xy' = \tan x

y' = \tan x

7.

y=(x2+1)5y = (x^2 + 1)^5

Step 1: Chain rule ddxu5=5u4u′\frac{d}{dx} u^5 = 5u^4 u', u=x2+1u = x^2 + 1, u′=2xu' = 2x.

y′=5(x2+1)4⋅2x=10x(x2+1)4y' = 5(x^2 + 1)^4 \cdot 2x = 10x (x^2 + 1)^4

y' = 10x(x^{2 + 1)^{4}}

8.

y=1x2+1y = \frac{1}{x^2 + 1}

Step 1: Quotient rule ddxuv=u′v−uv′v2\frac{d}{dx} \frac{u}{v} = \frac{u'v - uv'}{v^2}, u=1u=1, v=x2+1v = x^2 + 1.

u′=0,v′=2xu' = 0, \quad v' = 2x

Step 2: Substitute.

y′=0⋅(x2+1)−1⋅2x(x2+1)2=−2x(x2+1)2y' = \frac{0 \cdot (x^2 + 1) - 1 \cdot 2x}{(x^2 + 1)^2} = \frac{-2x}{(x^2 + 1)^2}

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

9.

y=[f(x)]−1−h(x)y = [f(x)]^{-1} - h(x)

Step 1: Chain rule for [f(x)]−1[f(x)]^{-1}: ddxu−1=−u−2u′\frac{d}{dx} u^{-1} = -u^{-2} u', u=f(x)u = f(x), u′=f′(x)u' = f'(x).

ddx[f(x)]−1=−[f(x)]−2f′(x)=−f′(x)[f(x)]2\frac{d}{dx} [f(x)]^{-1} = -[f(x)]^{-2} f'(x) = -\frac{f'(x)}{[f(x)]^2}

Step 2: ddx[−h(x)]=−h′(x)\frac{d}{dx} [-h(x)] = -h'(x).

Step 3: Sum.

y′=−f′(x)[f(x)]2−h′(x)y' = -\frac{f'(x)}{[f(x)]^2} - h'(x)

y' = -\dfrac{f'(x){[f(x)]^{2}} - h'(x)}

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Quick Answer

1. y = x^(1)/(2) - 3x^(3)/(4) Step 1: Power rule (d)/(dx) x^n = n x^n-1.

Differentiate the following with respect to x: 1. y = x^(1/2) + 1/x2 - 3/x3 + 4x^(1/4)
Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

ScanSolve AI Answer
1. y = x^(1)/(2) - 3x^(3)/(4) Step 1: Power rule (d)/(dx) x^n = n x^n-1. (d)/(dx) ( x^(1)/(2) ) = (1)/(2) x^(1)/(2)-1 = (1)/(2) x^-(1)/(2) Step 2: (d)/(dx) ( -3x^(3)/(4) ) = -3 · (3)/(4) x^(3)/(4)-1 = -(9)/(4) x^-(1)/(4) Step 3: Sum the derivatives. y' = (1)/(2) x^-(1)/(2) - (9)/(4) x^-(1)/(4) y' = (1)/(2sqrt(x)) - (9)/(4 x^1/4) 2. y = x - x - 2 x Step 1: (d)/(dx) x = 1, (d)/(dx) (- x) = x, (d)/(dx) (-2 x) = -2 x. Step 2: Sum the derivatives. y' = 1 + x - 2 x y' = 1 + x - 2 x 3. y = e^x x Step 1: Product rule (d)/(dx) (uv) = u'v + uv', where u = e^x, v = x. u' = e^x, v' = ^2 x Step 2: Substitute. y' = e^x x + e^x ^2 x Step 3: Factor. y' = e^x ( x + ^2 x) y' = e^x ( x + ^2 x) 4. y = ( x)^2 - 4 x Step 1: Chain rule for ( x)^2: let u = x, (dy)/(du) = 2u, (du)/(dx) = (1)/(x). (d)/(dx) ( x)^2 = 2 x · (1)/(x) Step 2: (d)/(dx) (-4 x) = -4 · (1)/(x). Step 3: Sum. y' = (2 x)/(x) - (4)/(x) = (2 x - 4)/(x) y' = (2 x - 4)/(x) 5. y = (3x + 1) Step 1: Chain rule (d)/(dx) u = u · u', u = 3x + 1, u' = 3. y' = (3x + 1) · 3 = 3 (3x + 1) y' = 3(3x + 1) 6. y = ( x) Step 1: Chain rule (d)/(dx) u = (1)/(u) u', u = x, u' = x x. y' = (1)/( x) · x x Step 2: Simplify. y' = x y' = x 7. y = (x^2 + 1)^5 Step 1: Chain rule (d)/(dx) u^5 = 5u^4 u', u = x^2 + 1, u' = 2x. y' = 5(x^2 + 1)^4 · 2x = 10x (x^2 + 1)^4 y' = 10x(x^2 + 1)^4 8. y = (1)/(x^2 + 1) Step 1: Quotient rule (d)/(dx) (u)/(v) = (u'v - uv')/(v^2), u=1, v = x^2 + 1. u' = 0, v' = 2x Step 2: Substitute. y' = (0 · (x^2 + 1) - 1 · 2x)/((x^2 + 1)^2) = (-2x)/((x^2 + 1)^2) y' = -(2x)/((x^2) + 1)^2 9. y = [f(x)]^-1 - h(x) Step 1: Chain rule for [f(x)]^-1: (d)/(dx) u^-1 = -u^-2 u', u = f(x), u' = f'(x). (d)/(dx) [f(x)]^-1 = -[f(x)]^-2 f'(x) = -(f'(x))/([f(x)]^2) Step 2: (d)/(dx) [-h(x)] = -h'(x). Step 3: Sum. y' = -(f'(x))/([f(x)]^2) - h'(x) y' = -(f'(x))/([f(x)]^2) - h'(x)