Math Problem Statement

Differenciate

Differentiate several functions including f(x) = x - 3 sin x, g(t) = t^3 cos t, y = sin 4x, and others.

Solution

The image presents several functions to differentiate. Let's go through each one systematically and find the derivatives.

(a) f(x)=x3sinxf(x) = x - 3 \sin x

The derivative is: f(x)=ddx(x)ddx(3sinx)=13cosxf'(x) = \frac{d}{dx} \left( x \right) - \frac{d}{dx} \left( 3 \sin x \right) = 1 - 3 \cos x

(b) g(t)=t3costg(t) = t^3 \cos t

Using the product rule: g(t)=ddt(t3)cost+t3ddt(cost)g'(t) = \frac{d}{dt} \left( t^3 \right) \cdot \cos t + t^3 \cdot \frac{d}{dt} \left( \cos t \right) g(t)=3t2costt3sintg'(t) = 3t^2 \cos t - t^3 \sin t

(c) f(x)=xsinxf(x) = x \sin x

Using the product rule: f(x)=ddx(x)sinx+xddx(sinx)f'(x) = \frac{d}{dx} \left( x \right) \cdot \sin x + x \cdot \frac{d}{dx} \left( \sin x \right) f(x)=sinx+xcosxf'(x) = \sin x + x \cos x

(d) y=xcosxsinxy = x \cos x \sin x

This is another product of functions, and we can treat cosxsinx\cos x \sin x as a single term for differentiation: y=ddx(x)cosxsinx+xddx(cosxsinx)y' = \frac{d}{dx} \left( x \right) \cdot \cos x \sin x + x \cdot \frac{d}{dx} \left( \cos x \sin x \right) Using the product rule for cosxsinx\cos x \sin x: ddx(cosxsinx)=sin2x+cos2x\frac{d}{dx} \left( \cos x \sin x \right) = -\sin^2 x + \cos^2 x Thus: y=cosxsinx+x(sin2x+cos2x)y' = \cos x \sin x + x (-\sin^2 x + \cos^2 x)

(e) y=1+sinxx+cosxy = \frac{1 + \sin x}{x + \cos x}

This is a quotient, so we use the quotient rule: y=(x+cosx)ddx(1+sinx)(1+sinx)ddx(x+cosx)(x+cosx)2y' = \frac{(x + \cos x) \cdot \frac{d}{dx}(1 + \sin x) - (1 + \sin x) \cdot \frac{d}{dx}(x + \cos x)}{(x + \cos x)^2} Simplifying each term: y=(x+cosx)cosx(1+sinx)(1sinx)(x+cosx)2y' = \frac{(x + \cos x) \cos x - (1 + \sin x) (1 - \sin x)}{(x + \cos x)^2}

(f) y=xcosxy = \frac{x}{\cos x}

This is another quotient, so we apply the quotient rule: y=cosx1x(sinx)cos2xy' = \frac{\cos x \cdot 1 - x (-\sin x)}{\cos^2 x} y=cosx+xsinxcos2xy' = \frac{\cos x + x \sin x}{\cos^2 x}

(g) y=sin4xy = \sin 4x

Using the chain rule: y=cos4x4=4cos4xy' = \cos 4x \cdot 4 = 4 \cos 4x

(h) y=cos(sinx)y = \cos(\sin x)

Using the chain rule: y=sin(sinx)cosxy' = -\sin(\sin x) \cdot \cos x

(i) y=sinxy = \sqrt{\sin x}

Rewrite as y=(sinx)1/2y = (\sin x)^{1/2}, then use the chain rule: y=12(sinx)1/2cosx=cosx2sinxy' = \frac{1}{2} (\sin x)^{-1/2} \cdot \cos x = \frac{\cos x}{2 \sqrt{\sin x}}

(j) y=cosxy = \cos \sqrt{x}

Using the chain rule: y=sinx12x=sinx2xy' = -\sin \sqrt{x} \cdot \frac{1}{2\sqrt{x}} = \frac{-\sin \sqrt{x}}{2\sqrt{x}}

(k) f(x)=sin5(3x)f(x) = \sin^5(3x)

This requires both the chain rule and the power rule: f(x)=5sin4(3x)cos(3x)3=15sin4(3x)cos(3x)f'(x) = 5 \sin^4(3x) \cdot \cos(3x) \cdot 3 = 15 \sin^4(3x) \cos(3x)

(l) y=4sin2x+5cos2xy = 4 \sin^2 x + 5 \cos^2 x

Differentiate each term separately: y=42sinxcosx+52cosx(sinx)=8sinxcosx10sinxcosxy' = 4 \cdot 2 \sin x \cos x + 5 \cdot 2 \cos x (-\sin x) = 8 \sin x \cos x - 10 \sin x \cos x y=2sinxcosxy' = -2 \sin x \cos x

Would you like any further clarifications on these derivatives?


5 Relative Questions:

  1. How does the product rule work in differentiating composite functions?
  2. Can you explain the chain rule used in y=sin4xy = \sin 4x and y=cosxy = \cos \sqrt{x}?
  3. What is the quotient rule and how is it applied to rational functions like in (e)?
  4. How would you differentiate a function with a trigonometric term raised to a power, as in (k)?
  5. What is the significance of simplifying the derivative after applying the quotient rule?

Tip:

When differentiating composite functions, always consider the order of operations to apply the chain rule or product rule accurately.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Product Rule
Quotient Rule
Chain Rule
Trigonometric Functions

Formulas

Product Rule: d(uv)/dx = u'v + uv'
Quotient Rule: d(u/v)/dx = (v u' - u v') / v^2
Chain Rule: d(f(g(x)))/dx = f'(g(x)) * g'(x)

Theorems

Basic Derivative Rules
Trigonometric Derivatives
Chain Rule
Product Rule
Quotient Rule

Suitable Grade Level

Undergraduate or Advanced High School (Calculus Level)