Math Problem Statement

Q5: Find dy/dx for the given functions: (a) y = (2x^4 - 5)^3 - 3x^2, (b) y = log_e(3x^4 + 2x - 3), (c) y = cos(x^2 + 1) * sin^3(x), (d) y = e^x / (1 - x^2)

Solution

The question requires calculating the derivatives dydx\frac{dy}{dx} for four different functions yy as shown in the image. Let’s go over each part and verify if the solutions are correct.


Part (a)

y=(2x45)33x2y = (2x^4 - 5)^3 - 3x^2 To find dydx\frac{dy}{dx}, we apply the chain rule and differentiate each term separately:

  1. Derivative of (2x45)3(2x^4 - 5)^3:
    Using the chain rule: ddx((2x45)3)=3(2x45)2ddx(2x45)\frac{d}{dx} \left( (2x^4 - 5)^3 \right) = 3(2x^4 - 5)^2 \cdot \frac{d}{dx}(2x^4 - 5) where ddx(2x45)=8x3\frac{d}{dx}(2x^4 - 5) = 8x^3 So, ddx((2x45)3)=3(2x45)28x3=24x3(2x45)2\frac{d}{dx} \left( (2x^4 - 5)^3 \right) = 3(2x^4 - 5)^2 \cdot 8x^3 = 24x^3 (2x^4 - 5)^2

  2. Derivative of 3x2-3x^2: ddx(3x2)=6x\frac{d}{dx}(-3x^2) = -6x

Combining these: dydx=24x3(2x45)26x\frac{dy}{dx} = 24x^3 (2x^4 - 5)^2 - 6x The solution provided matches this answer.


Part (b)

y=ln(3x4+2x3)y = \ln(3x^4 + 2x - 3) To differentiate, we use the chain rule with the natural logarithm:

  1. Derivative of ln(3x4+2x3)\ln(3x^4 + 2x - 3): dydx=13x4+2x3ddx(3x4+2x3)\frac{dy}{dx} = \frac{1}{3x^4 + 2x - 3} \cdot \frac{d}{dx}(3x^4 + 2x - 3) where ddx(3x4+2x3)=12x3+2\frac{d}{dx}(3x^4 + 2x - 3) = 12x^3 + 2 So, dydx=12x3+23x4+2x3\frac{dy}{dx} = \frac{12x^3 + 2}{3x^4 + 2x - 3}

The provided solution also matches this answer.


Part (c)

y=cos(x2+1)sin3(x)y = \cos(x^2 + 1) \sin^3(x) To differentiate, we use the product rule, letting:

  • u=cos(x2+1)u = \cos(x^2 + 1)
  • v=sin3(x)v = \sin^3(x)
  1. Derivative of u=cos(x2+1)u = \cos(x^2 + 1): dudx=sin(x2+1)2x=2xsin(x2+1)\frac{du}{dx} = -\sin(x^2 + 1) \cdot 2x = -2x \sin(x^2 + 1)

  2. Derivative of v=sin3(x)v = \sin^3(x): dvdx=3sin2(x)cos(x)=3sin2(x)cos(x)\frac{dv}{dx} = 3 \sin^2(x) \cdot \cos(x) = 3 \sin^2(x) \cos(x)

Using the product rule: dydx=udvdx+vdudx\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} =cos(x2+1)3sin2(x)cos(x)sin3(x)2xsin(x2+1)= \cos(x^2 + 1) \cdot 3 \sin^2(x) \cos(x) - \sin^3(x) \cdot 2x \sin(x^2 + 1) =3cos(x2+1)sin2(x)cos(x)2xsin3(x)sin(x2+1)= 3 \cos(x^2 + 1) \sin^2(x) \cos(x) - 2x \sin^3(x) \sin(x^2 + 1)


Part (d)

y=ex1x2y = \frac{e^x}{1 - x^2} To differentiate, we use the quotient rule, letting:

  • u=exu = e^x
  • v=1x2v = 1 - x^2
  1. Derivative of u=exu = e^x: dudx=ex\frac{du}{dx} = e^x

  2. Derivative of v=1x2v = 1 - x^2: dvdx=2x\frac{dv}{dx} = -2x

Using the quotient rule: dydx=vdudxudvdxv2\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} =(1x2)exex(2x)(1x2)2= \frac{(1 - x^2)e^x - e^x(-2x)}{(1 - x^2)^2} =ex(1x2+2x)(1x2)2= \frac{e^x(1 - x^2 + 2x)}{(1 - x^2)^2} This matches the answer given.


Would you like further details on any specific part?

Related Questions:

  1. How is the chain rule applied to more complex functions?
  2. What are some advanced uses of the product rule in calculus?
  3. How does the quotient rule differ from other differentiation rules?
  4. Can logarithmic differentiation simplify problems like part (b)?
  5. How does trigonometric differentiation apply in physics and engineering?

Tip:

When working with derivatives, identifying the type of function (product, quotient, composite) helps in selecting the correct differentiation rule.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Chain Rule
Product Rule
Quotient Rule
Logarithmic Differentiation

Formulas

Chain rule: d/dx[f(g(x))] = f'(g(x)) * g'(x)
Product rule: d/dx[u*v] = u'v + uv'
Quotient rule: d/dx[u/v] = (v*u' - u*v') / v^2
Logarithmic differentiation: d/dx[ln(f(x))] = f'(x) / f(x)

Theorems

Chain Rule
Product Rule
Quotient Rule

Suitable Grade Level

Grades 11-12