text: # Thread: calculus with exponents
  
  1. ## calculus with exponents
  
  I can't get a clear graph on here, but you can graph it on a graphing calculator, the graph of the curve with equation: y=e^(2x)cosx
  On this graph, P(a,b) is labelled in the first quadrant as the graph is increasing.
  
  a) Show that the first derivative=e^(2x)(2cosx-sinx)
  b) Find second derivative
  
  At the point P(a,b), the second derivative = 0.
  c) Use results from a) and b) to prove that
  i) tan a = 3/4
  ii) the gradient of the curve at P is e^(2a)
  
  Ok, so I can do a) and b), the second derivative is e^(2x)(3cosx-4sinx) and this is correct. I can prove c)i), but I dont know how to prove ii).
  
  If I substitute 'a' into the equation of the derivative, then it's: e^(2a)(2cosa-sina), how is the answer e^(2a)???
  
  2. Originally Posted by bhuang
  I can't get a clear graph on here, but you can graph it on a graphing calculator, the graph of the curve with equation: y=e^(2x)cosx
  On this graph, P(a,b) is labelled in the first quadrant as the graph is increasing.
  
  a) Show that the first derivative=e^(2x)(2cosx-sinx)
  b) Find second derivative
  
  At the point P(a,b), the second derivative = 0.
  c) Use results from a) and b) to prove that
  i) tan a = 3/4
  ii) the gradient of the curve at P is e^(2a)
  
  Ok, so I can do a) and b), the second derivative is e^(2x)(3cosx-4sinx) and this is correct. I can prove c)i), but I dont know how to prove ii).
  
  If I substitute 'a' into the equation of the derivative, then it's: e^(2a)(2cosa-sina), how is the answer e^(2a)???
  you know $\tan{a} = \frac{3}{4}$ ... so, what would be the values of $\sin{a}$ and $\cos{a}$ ? ... and how will those values determine the value of $e^{2a}(2\cos{a}-\sin{a})$ ?
explanation: The text extract is a discussion thread about a calculus problem involving a function with exponents and trigonometric functions. The user presents a problem with multiple parts and provides their progress on solving it. The user is specifically stuck on part c)ii) of the problem, where they need to prove that the gradient of the curve at point P is \( e^{2a} \).
  
  The thread includes a question and a partial answer, but the user is seeking further clarification on a specific part of the solution.
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: # Thread: calculus with exponents
  
  1. ## calculus with exponents
  
  I can't get a clear graph on here, but you can graph it on a graphing calculator, the graph of the curve with equation: y=e^(2x)cosx
  On this graph, P(a,b) is labelled in the first quadrant as the graph is increasing.
  
  a) Show that the first derivative=e^(2x)(2cosx-sinx)
  b) Find second derivative
  
  At the point P(a,b), the second derivative = 0.
  c) Use results from a) and b) to prove that
  i) tan a = 3/4
  ii) the gradient of the curve at P is e^(2a)
  
  Ok, so I can do a) and b), the second derivative is e^(2x)(3cosx-4sinx) and this is correct. I can prove c)i), but I dont know how to prove ii).
  
  If I substitute 'a' into the equation of the derivative, then it's: e^(2a)(2cosa-sina), how is the answer e^(2a)???
  
  2. Originally Posted by bhuang
  I can't get a clear graph on here, but you can graph it on a graphing calculator, the graph of the curve with equation: y=e^(2x)cosx
  On this graph, P(a,b) is labelled in the first quadrant as the graph is increasing.
  
  a) Show that the first derivative=e^(2x)(2cosx-sinx)
  b) Find second derivative
  
  At the point P(a,b), the second derivative = 0.
  c) Use results from a) and b) to prove that
  i) tan a = 3/4
  ii) the gradient of the curve at P is e^(2a)
  
  Ok, so I can do a) and b), the second derivative is e^(2x)(3cosx-4sinx) and this is correct. I can prove c)i), but I dont know how to prove ii).
  
  If I substitute 'a' into the equation of the derivative, then it's: e^(2a)(2cosa-sina), how is the answer e^(2a)???
  you know $\tan{a} = \frac{3}{4}$ ... so, what would be the values of $\sin{a}$ and $\cos{a}$ ? ... and how will those values determine the value of $e^{2a}(2\cos{a}-\sin{a})$ ?
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The text extract is a discussion thread about a calculus problem involving a function with exponents and trigonometric functions. The user presents a problem with multiple parts and provides their progress on solving it. The user is specifically stuck on part c)ii) of the problem, where they need to prove that the gradient of the curve at point P is \( e^{2a} \).
  
  The thread includes a question and a partial answer, but the user is seeking further clarification on a specific part of the solution.
  
  Label: question-and-answer