- Differentiate functions (power, product, quotient, chain rule) with full step breakdowns
- Evaluate definite and indefinite integrals, including u-substitution and integration by parts
- Compute one-sided and two-sided limits, including L'Hôpital's Rule cases
- Expand Taylor and Maclaurin series to a chosen number of terms
- Solve related rates and optimization problems with annotated setup steps
- Graph the function alongside its derivative or integral for visual context
- Choose the operation: derivative, integral, limit, or series
- Type your function using standard notation (e.g., x^2, sin(x), e^x, ln(x))
- For integrals, specify bounds if you need a definite integral; leave blank for indefinite
- For limits, enter the point the variable approaches (including infinity)
- Click Solve to see the step-by-step breakdown and final simplified answer
Step-by-step solution
1
Identify outer function: u^4, inner function: u = 3x^2 + 5x
2
Differentiate outer function: d/du [u^4] = 4u^3
3
Differentiate inner function: du/dx = 6x + 5
4
Apply chain rule: f'(x) = 4u^3 * du/dx
5
Substitute back: f'(x) = 4(3x^2 + 5x)^3 * (6x + 5)
6
Final answer: f'(x) = 4(6x + 5)(3x^2 + 5x)^3
Step-by-step solution
1
Let u = x^2, so du = 2x dx (this matches the 2x term exactly)
2
Rewrite bounds: when x = 0, u = 0; when x = sqrt(pi/2), u = pi/2
3
Substitute: integral becomes integral of cos(u) du from 0 to pi/2
4
Integrate: antiderivative of cos(u) is sin(u)
5
Evaluate at bounds: sin(pi/2) - sin(0) = 1 - 0
6
Final answer: 1
Step-by-step solution
1
Check direct substitution: gives 0/0, an indeterminate form
2
Apply L'Hopital's Rule (1st time): differentiate top and bottom
3
New expression: (cos(x) - 1) / (3x^2), still 0/0 at x = 0
4
Apply L'Hopital's Rule (2nd time): (-sin(x)) / (6x), still 0/0
5
Apply L'Hopital's Rule (3rd time): (-cos(x)) / 6
6
Substitute x = 0: -cos(0)/6 = -1/6
7
Final answer: -1/6
| Rule | Formula | Used For |
|---|---|---|
| Power Rule | d/dx[x^n] = n·x^(n-1) | Polynomial derivatives |
| Product Rule | d/dx[uv] = u'v + uv' | Derivative of two multiplied functions |
| Quotient Rule | d/dx[u/v] = (u'v - uv')/v^2 | Derivative of a fraction of functions |
| Chain Rule | d/dx[f(g(x))] = f'(g(x))·g'(x) | Composite functions |
| Power Rule for Integrals | ∫x^n dx = x^(n+1)/(n+1) + C, n ≠ -1 | Basic polynomial integrals |
| Integration by Parts | ∫u dv = uv - ∫v du | Products of functions in integrals |
Q1Does this calculus solver show full steps, not just the final answer?
Yes. Every result includes the rule applied at each stage (power rule, chain rule, substitution, etc.) so you can follow the logic or check homework line by line.
Q2Can it handle both definite and indefinite integrals?
Yes. Leave the bounds blank for an indefinite integral (with a +C constant), or enter lower and upper bounds for a definite integral with a numeric result.
Q3Does it support multivariable calculus?
The core tool focuses on single-variable derivatives, integrals, limits, and series. Partial derivatives and multiple integrals are handled on a separate multivariable calculus solver page.
Q4Is this calculus solver free to use?
Yes, solving problems and viewing step-by-step solutions is free with no account required.
Q5Can I use this to check my calculus homework?
Yes, many students use it to verify hand-worked derivatives and integrals, but it's best used to check understanding rather than replace practicing the steps yourself.