Derivative Calculator

Intro: MathGPT.today shows each differentiation step and names the rule used. You can then generate follow-up practice questions.

Worked example

• Differentiate: $\dfrac{d}{dx}\,(x^3 - 4x + 6)$



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• Split (linearity): $\dfrac{d}{dx}(x^3 - 4x + 6)=\dfrac{d}{dx}(x^3)-\dfrac{d}{dx}(4x)+\dfrac{d}{dx}(6)$



• Power rule: $\dfrac{d}{dx}(x^3)=3x^2$



• Linear term: $\dfrac{d}{dx}(4x)=4$ so contribution is $-4$



• Constant rule: $\dfrac{d}{dx}(6)=0$



• Combine: $\boxed{3x^2-4}$





• Differentiate: $\;y=(x^2 e^{3x})\,\ln(x^2+1)\; +\; \dfrac{\sin(2x)}{x^3}$



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• Goal: Find $y'\,$ for $\;y=(x^2 e^{3x})\,\ln(x^2+1) + \dfrac{\sin(2x)}{x^3}$. We'll use Product, Chain, and Quotient rules.



• Split by linearity: $y' = \dfrac{d}{dx}\big[(x^2 e^{3x})\,\ln(x^2+1)\big] \, + \, \dfrac{d}{dx}\Big[\dfrac{\sin(2x)}{x^3}\Big]$.



• — Product block —



• Let $A=x^2 e^{3x}$ and $B=\ln(x^2+1)$. Then $(AB)'=A'B + AB'$. We compute $A'$ and $B'$ separately.



• 1) Derivative of $A=x^2 e^{3x}$ (product again):



• Let $u=x^2$ and $v=e^{3x}$. Then $A'=u'v + uv'$.



• • $u' = 2x$.



• • $v' = e^{3x}\cdot (3)$ by Chain rule (since $(e^{t})'=e^{t}$ and $t=3x$ gives $t'=3$).



• Therefore $A' = (2x)e^{3x} + x^2(3e^{3x}) = e^{3x}(2x + 3x^2)$.



• 2) Derivative of $B=\ln(x^2+1)$ (chain):



• $(\ln g(x))' = \dfrac{g'(x)}{g(x)}$ with $g(x)=x^2+1$.



• $g'(x)=2x$, so $B' = \dfrac{2x}{x^2+1}$.



• Assemble the product: $(AB)' = A'B + AB' = e^{3x}(2x+3x^2)\,\ln(x^2+1) + x^2 e^{3x}\cdot \dfrac{2x}{x^2+1}$.



• — Quotient block —



• Now $C(x)=\dfrac{\sin(2x)}{x^3}$. Use Quotient rule: $\big(\dfrac{u}{v}\big)' = \dfrac{u'v - u v'}{v^2}$.



• Take $u=\sin(2x)$ and $v=x^3$.



• • $u' = \cos(2x)\cdot 2 = 2\cos(2x)$ by Chain rule.



• • $v' = 3x^2$.



• Thus $C'(x) = \dfrac{(2\cos(2x))x^3 - \sin(2x)\cdot (3x^2)}{x^6}$.



• Simplify $C'(x)$ by factoring $x^2$ in the numerator and canceling with $x^6$: $C'(x)=\dfrac{2x\cos(2x) - 3\sin(2x)}{x^4}$.



• — Combine everything —



• Therefore



• $$\begin{aligned} y' &= e^{3x}(2x+3x^2)\,\ln(x^2+1) \; + \; x^2 e^{3x}\cdot \dfrac{2x}{x^2+1} \; + \; \dfrac{2x\cos(2x) - 3\sin(2x)}{x^4}. \end{aligned}$$



• — Optional tidy factorization —



• Group the first two terms: pull out $e^{3x}$:



• $$y' = e^{3x}\Big[(2x+3x^2)\,\ln(x^2+1) \; + \; \dfrac{2x^3}{x^2+1}\Big] \; + \; \dfrac{2x\cos(2x) - 3\sin(2x)}{x^4}.$$



• Final result:



• $$\boxed{\;y' = e^{3x}\left((2x+3x^2)\,\ln(x^2+1) + \dfrac{2x^3}{x^2+1}\right) + \dfrac{2x\cos(2x) - 3\sin(2x)}{x^4}\;}$$

Why choose Mathix?

• Get clear, step-by-step solutions that explain the “why,” not just the answer.
• See the rules used at each step (power, product, quotient, chain, and more).
• Optional animated walk-throughs to make tricky ideas click faster.
• Clean LaTeX rendering for notes, homework, and study guides.

Differentiate x^3 - 4x + 6

• d/dx (x^3) = 3x^2


• d/dx (-4x) = -4


• d/dx (6) = 0


• Result: 3x^2 - 4

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