Math Problem Statement

solve these maths

Solve these math problems.

Solution

Let us solve each of the problems one by one.


1. Find y(x)y'(x), where:

y(x)=8+4x2+3x4151x3.y(x) = \frac{8 + 4x^2 + 3x^4}{15} \cdot \sqrt[3]{1 - x}.

Solution:

We apply the product rule and chain rule: Let: u(x)=8+4x2+3x415,v(x)=1x3.u(x) = \frac{8 + 4x^2 + 3x^4}{15}, \quad v(x) = \sqrt[3]{1 - x}.

Derivatives:

  1. u(x)=ddx(8+4x2+3x415)=115(8x+12x3)u'(x) = \frac{d}{dx} \left( \frac{8 + 4x^2 + 3x^4}{15} \right) = \frac{1}{15}(8x + 12x^3).
  2. v(x)=ddx((1x)1/3)=13(1x)2/3v'(x) = \frac{d}{dx} \left( (1 - x)^{1/3} \right) = \frac{-1}{3}(1 - x)^{-2/3}.

Product Rule:

y(x)=u(x)v(x)+u(x)v(x).y'(x) = u'(x)v(x) + u(x)v'(x).

Substitute u(x),u(x),v(x),v(x)u(x), u'(x), v(x), v'(x) to find y(x)y'(x).


2. Find y(x)y'(x), if:

x=t2+3t,y=t4+2t3.x = t^2 + 3t, \quad y = t^4 + 2t^3.

Solution:

We find dydx\frac{dy}{dx} using the chain rule: dydx=dydtdxdt.\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}.

Derivatives:

  1. dxdt=2t+3\frac{dx}{dt} = 2t + 3.
  2. dydt=4t3+6t2\frac{dy}{dt} = 4t^3 + 6t^2.

Substitute to get: dydx=4t3+6t22t+3.\frac{dy}{dx} = \frac{4t^3 + 6t^2}{2t + 3}.


3. Find y(x)y''(x), if:

y=(1+x2)arctan(x).y = (1 + x^2) \arctan(x).

Solution:

We use the product rule and then differentiate again: y=ddx((1+x2)arctan(x))=arctan(x)2x+(1+x2)11+x2.y' = \frac{d}{dx} \left( (1 + x^2) \arctan(x) \right) = \arctan(x) \cdot 2x + (1 + x^2) \cdot \frac{1}{1 + x^2}.

Simplify yy', and then differentiate again to get yy''.


4. Find sin(x)cos(x)+4dx\int \frac{\sin(x)}{\cos(x) + 4} \, dx.

Solution:

We use substitution: Let: u=cos(x)+4,dudx=sin(x).u = \cos(x) + 4, \quad \frac{du}{dx} = -\sin(x).

Then: sin(x)cos(x)+4dx=1udu=lnu+C=lncos(x)+4+C.\int \frac{\sin(x)}{\cos(x) + 4} \, dx = -\int \frac{1}{u} \, du = -\ln|u| + C = -\ln|\cos(x) + 4| + C.


5. Find x2(lnx+10)dx\int x^2 (\ln x + 10) \, dx.

Solution:

We split into two parts: x2(lnx+10)dx=x2lnxdx+10x2dx.\int x^2 (\ln x + 10) \, dx = \int x^2 \ln x \, dx + \int 10x^2 \, dx.

  1. For x2lnxdx\int x^2 \ln x \, dx, use integration by parts: Let u=lnx,dv=x2dxu = \ln x, \, dv = x^2 dx. Then du=1xdx,v=x33du = \frac{1}{x} dx, \, v = \frac{x^3}{3}. x2lnxdx=x33lnxx331xdx=x33lnxx39.\int x^2 \ln x \, dx = \frac{x^3}{3} \ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx = \frac{x^3}{3} \ln x - \frac{x^3}{9}.

  2. For 10x2dx\int 10x^2 \, dx: 10x2dx=10x33.\int 10x^2 \, dx = \frac{10x^3}{3}.

Combine the results: x2(lnx+10)dx=x33lnxx39+10x33+C.\int x^2 (\ln x + 10) \, dx = \frac{x^3}{3} \ln x - \frac{x^3}{9} + \frac{10x^3}{3} + C.


Would you like detailed steps for any of these solutions or further clarifications? Here are 5 related questions you can explore:

  1. How do you apply the product rule for derivatives in question 1?
  2. Can we explore the chain rule for parametric differentiation in question 2?
  3. How do we simplify higher-order derivatives, like in question 3?
  4. What alternative substitutions could be used for integrals like in question 4?
  5. How do you simplify results after applying integration by parts?

Tip: Always organize your work when solving multi-step derivatives or integrals to avoid errors!

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Math Problem Analysis

Mathematical Concepts

Differentiation
Chain Rule
Product Rule
Integration
Parametric Differentiation
Integration by Parts
Substitution Method

Formulas

Product Rule: (uv)' = u'v + uv'
Chain Rule: dy/dx = (dy/dt) / (dx/dt)
Derivative of arctan(x): d/dx[arctan(x)] = 1 / (1 + x^2)
Integration by Parts: ∫u dv = uv - ∫v du
Substitution for integrals

Theorems

Chain Rule
Fundamental Theorem of Calculus

Suitable Grade Level

Undergraduate Mathematics