MarkFL said:Since you posted in the Pre-Calculus forum, I suppose you are to use the definition:
\(\displaystyle f'(x)\equiv\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\)
So, for $f(x)=\sin(x)$, we have:
\(\displaystyle f'(x)\equiv\lim_{h\to0}\frac{\sin(x+h)-\sin(x)}{h}\)
Using the angle-sum identity for the sine function, we may write:
\(\displaystyle f'(x)=\lim_{h\to0}\frac{\sin(x)(\cos(h)-1)+\cos(x)\sin(h)}{h}\)
\(\displaystyle f'(x)=\sin(x)\lim_{h\to0}\frac{\cos(h)-1}{h}+\cos(x)\lim_{h\to0}\frac{\sin(h)}{h}\)
\(\displaystyle f'(x)=\sin(x)\lim_{h\to0}\frac{\cos^2(h)-1}{h(\cos(h)+1)}+\cos(x)\lim_{h\to0}\frac{\sin(h)}{h}\)
Using a Pythagorean identity on the first limit on the RHS, we obtain:
\(\displaystyle f'(x)=-\sin(x)\lim_{h\to0}\frac{\sin^2(h)}{h(\cos(h)+1)}+\cos(x)\lim_{h\to0}\frac{\sin(h)}{h}\)
\(\displaystyle f'(x)=-\sin(x)\lim_{h\to0}\frac{\sin(h)}{h}\cdot\lim_{h\to0}\frac{\sin(h)}{(\cos(h)+1)}+\cos(x)\lim_{h\to0}\frac{\sin(h)}{h}\)
Using the fact that \(\displaystyle \lim_{u\to0}\frac{\sin(u)}{u}=1\), we have:
\(\displaystyle f'(x)=-\sin(x)(1)(0)+\cos(x)(1)=\cos(x)\)
Can you now obtain a result for $f(x)=\cos(x)$?
rahulk said:Hi
Thanks for your suggestion
But what about c=pi/4 where are used
c=pi/4 in the answer
MarkFL said:Since you posted in the Pre-Calculus forum, I suppose you are to use the definition:
\(\displaystyle f'(x)\equiv\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\)
So, for $f(x)=\sin(x)$, we have:
\(\displaystyle f'(x)\equiv\lim_{h\to0}\frac{\sin(x+h)-\sin(x)}{h}\)
Using the angle-sum identity for the sine function, we may write:
\(\displaystyle f'(x)=\lim_{h\to0}\frac{\sin(x)(\cos(h)-1)+\cos(x)\sin(h)}{h}\)
\(\displaystyle f'(x)=\sin(x)\lim_{h\to0}\frac{\cos(h)-1}{h}+\cos(x)\lim_{h\to0}\frac{\sin(h)}{h}\)
\(\displaystyle f'(x)=\sin(x)\lim_{h\to0}\frac{\cos^2(h)-1}{h(\cos(h)+1)}+\cos(x)\lim_{h\to0}\frac{\sin(h)}{h}\)
Using a Pythagorean identity on the first limit on the RHS, we obtain:
\(\displaystyle f'(x)=-\sin(x)\lim_{h\to0}\frac{\sin^2(h)}{h(\cos(h)+1)}+\cos(x)\lim_{h\to0}\frac{\sin(h)}{h}\)
\(\displaystyle f'(x)=-\sin(x)\lim_{h\to0}\frac{\sin(h)}{h}\cdot\lim_{h\to0}\frac{\sin(h)}{(\cos(h)+1)}+\cos(x)\lim_{h\to0}\frac{\sin(h)}{h}\)
Using the fact that \(\displaystyle \lim_{u\to0}\frac{\sin(u)}{u}=1\), we have:
\(\displaystyle f'(x)=-\sin(x)(1)(0)+\cos(x)(1)=\cos(x)\)
Can you now obtain a result for $f(x)=\cos(x)$?
rahulk said:How sin(h)/(cos(h)+1 =0 please descibe
The derivative of a function f(x) at a point c is defined as the limit of the difference quotient as the interval between the two points approaches 0. In other words, it measures the rate of change of the function at a specific point.
To find the value of f'(x) at c, you can use the definition of the derivative and take the limit of the difference quotient. Alternatively, you can use the power rule, product rule, quotient rule, or chain rule depending on the form of the function.
The derivative is an important concept in calculus as it allows us to analyze the rate of change of a function and determine its maximum and minimum points. It also has various applications in physics, engineering, and economics.
No, the derivative can only be found at points where the function is continuous. If the function has a discontinuity at a certain point, the derivative at that point does not exist.
Yes, there are various rules and formulas such as the power rule, product rule, quotient rule, and chain rule that can help you find the derivative of a function at a specific point. However, it is important to understand the concept of the derivative and how these rules are derived.