text: # How to calculate $f'(0)$ and $f'$ $(\sqrt{2})$ while $f(x)$= $\int_{x}^{x^3} e^{t^{2}}dt\$?
  
  How to calculate $f'(0)$ and $f'$ $(\sqrt{2})$ while $f(x)$= $\int_{x}^{x^3} e^{t^{2}}dt\$?
  
  I thought about using the fundamental theorem of calculus, but im not sure im fully aware of how to use it in this case.
  
  any kind of help/directing, would be appreciated.
  
  Yes, you use the fundamental theorem. Let $F(x) = \int_a^x e^{t^2}\ dt$. Then
  
  $$f(x) = F(x^3)-F(x) = \int_x^{x^3} e^{t^2} \ dt$$
  
  So
  
  $$f'(x) = F'(x^3)\cdot 3x^2 - F'(x)$$
  
  where by the fundamental theorem,
  
  $$F'(x) = e^{x^2}$$
  
  • Lets suggest that $f(x)$ = $\int_{0}^{x^2-3x} (2 + sint^2) dt$ , what function would u suggest at the beginning for $F(x)$? Apr 13, 2015 at 22:51
  • It's the same idea, $F(x)=\int_a^x (2+\sin t^2)\ dt$. Then $f(x) = F(x^2-3x)-F(0)$. Or, you could just pick $a=0$ and use $f(x)=F(x^2-3x)$. Apr 14, 2015 at 1:19
  
  Some hints. The Fundamental Theorem says that $$\frac{\mathrm{d}}{\mathrm{d}u}\int_0^ue^{t^2}\,\mathrm{d}t=e^{u^2}\tag{1}$$ Let $u=x^3$, then \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\int_0^ue^{t^2}\,\mathrm{d}t &=\frac{\mathrm{d}u}{\mathrm{d}x}\frac{\mathrm{d}}{\mathrm{d}u}\int_0^ue^{t^2}\,\mathrm{d}t\\ &=3x^2\frac{\mathrm{d}}{\mathrm{d}u}\int_0^ue^{t^2}\,\mathrm{d}t\tag{2} \end{align} Then note that $$\int_x^ue^{t^2}\,\mathrm{d}t=\int_0^ue^{t^2}\,\mathrm{d}t-\int_0^xe^{t^2}\,\mathrm{d}t\tag{3}$$
explanation: The document contains a math question and a detailed explanation of how to solve it, including a step-by-step solution and additional hints. The question asks how to calculate \( f'(0) \) and \( f'(\sqrt{2}) \) for the function \( f(x) = \int_{x}^{x^3} e^{t^2} \, dt \).
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # How to calculate $f'(0)$ and $f'$ $(\sqrt{2})$ while $f(x)$= $\int_{x}^{x^3} e^{t^{2}}dt\$?
  
  How to calculate $f'(0)$ and $f'$ $(\sqrt{2})$ while $f(x)$= $\int_{x}^{x^3} e^{t^{2}}dt\$?
  
  I thought about using the fundamental theorem of calculus, but im not sure im fully aware of how to use it in this case.
  
  any kind of help/directing, would be appreciated.
  
  Yes, you use the fundamental theorem. Let $F(x) = \int_a^x e^{t^2}\ dt$. Then
  
  $$f(x) = F(x^3)-F(x) = \int_x^{x^3} e^{t^2} \ dt$$
  
  So
  
  $$f'(x) = F'(x^3)\cdot 3x^2 - F'(x)$$
  
  where by the fundamental theorem,
  
  $$F'(x) = e^{x^2}$$
  
  • Lets suggest that $f(x)$ = $\int_{0}^{x^2-3x} (2 + sint^2) dt$ , what function would u suggest at the beginning for $F(x)$? Apr 13, 2015 at 22:51
  • It's the same idea, $F(x)=\int_a^x (2+\sin t^2)\ dt$. Then $f(x) = F(x^2-3x)-F(0)$. Or, you could just pick $a=0$ and use $f(x)=F(x^2-3x)$. Apr 14, 2015 at 1:19
  
  Some hints. The Fundamental Theorem says that $$\frac{\mathrm{d}}{\mathrm{d}u}\int_0^ue^{t^2}\,\mathrm{d}t=e^{u^2}\tag{1}$$ Let $u=x^3$, then \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\int_0^ue^{t^2}\,\mathrm{d}t &=\frac{\mathrm{d}u}{\mathrm{d}x}\frac{\mathrm{d}}{\mathrm{d}u}\int_0^ue^{t^2}\,\mathrm{d}t\\ &=3x^2\frac{\mathrm{d}}{\mathrm{d}u}\int_0^ue^{t^2}\,\mathrm{d}t\tag{2} \end{align} Then note that $$\int_x^ue^{t^2}\,\mathrm{d}t=\int_0^ue^{t^2}\,\mathrm{d}t-\int_0^xe^{t^2}\,\mathrm{d}t\tag{3}$$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question and a detailed explanation of how to solve it, including a step-by-step solution and additional hints. The question asks how to calculate \( f'(0) \) and \( f'(\sqrt{2}) \) for the function \( f(x) = \int_{x}^{x^3} e^{t^2} \, dt \).
  
  Label: question-and-answer