Question

find the critical points of the function f(x)=3^(1/2) sinx+cosx (0,2pi)

Answer 1

take the derivative and then set equal to zero, solve for x

Answer 2

Is this your function?\[
f(x) = \sqrt 3\sin x+\cos x
\]

Answer 3

Is my statement correct @wio ?

Answer 4

In this case, that will get you the critical points.

Answer 5

thanks guys, im just having problem on converting it into its derivative

Answer 6

does your function look like the one @wio pointed out? If so I suggest you take the square root and turn it into a power

Answer 7

take that last part back...

Answer 8

pretty much yea

Answer 9

what is the derivative of sinx and cosx?

Answer 10

Want me to post some derivative rules?

Answer 11

yes please

Answer 12

\[
\large \text{Differentiation Rules}
\]Let \(c\) be constant with respect to \(x\).
Let \(u\) and \(v\) vary with respect to \(x\).
\[
\begin{array}{r|ccc|l}
1&d(c) &=& 0&\text{Constants}\\ \
2&d(x) &=& dx&\text{The variable of differentiation}\\
3&d(|x|) &=& \frac{x}{|x|}~dx&\text{Absolute value function}\\
4&d(x^c) &=& cx^{c-1}~dx&\text{Power functions}\\
5&d(\sqrt x) &=& \frac{dx}{2\sqrt x}&\text{Square root function}\\
6&d(c^x) &=& c^{x}\ln(c)~dx&\text{Exponential functions}\\
7&d(e^x) &=& e^x~dx&\text{Exponential function}\\
8&d(\log_c(x)) &=& \frac{dx}{x\ln(c)}&\text{Logarithmic functions}\\
9&d(\ln(x)) &=& \frac{dx}{x}&\text{Natural logarithmic functions}\\
\hline
10&d(cu) &=& c~du&\text{Constant times a function}\\
11&d(u+v) &=& du+dv&\text{Sum of functions}\\
12&d(u\cdot v) &=& v~du+u~dv&\text{Product of functions}\\
13&d\left(\frac uv\right) &=& \frac{v~du-u~dv}{v^2}&\text{Quotient of functions}\\
14&d(u\circ v) &=& \frac{du}{dv}~dv&\text{Composition of functions}\\
\end{array}
\]\[
\large \text{Trigonometric Differentiation Rules}
\]\[
\begin{array}{r|ccc|l}
1&d(\sin(x)) &=& \cos(x)~dx &\text{Sine}\\
2&d(\cos(x)) &=& -\sin(x)~dx &\text{Cosine}\\
3&d(\tan(x)) &=& \sec^2(x)~dx &\text{Tangent}\\
4&d(\cot(x)) &=& -\csc^2(x)~dx &\text{Cotangent}\\
5&d(\sec(x)) &=& \sec(x)\tan(x)~dx &\text{Secant}\\
6&d(\csc(x)) &=& -\csc(x)\cot(x)~dx &\text{Cosecant}\\
7&d(\sin^{-1}(x)) &=& \frac{dx}{\sqrt{1-x^2}} &\text{Arcsine}\\
8&d(\cos^{-1}(x)) &=& -\frac{dx}{\sqrt{1-x^2}} &\text{Arccosine}\\
9&d(\tan^{-1}(x)) &=& \frac{dx}{1+x^2} &\text{Arctangent}\\
10&d(\cot^{-1}(x)) &=& -\frac{dx}{1+x^2} &\text{Arccotangent}\\
\hline
11&d(\sinh(x)) &=& \cosh(x)~dx &\text{Hyperbolic Sine}\\
12&d(\cosh(x)) &=& \sinh(x)~dx &\text{Hyperbolic Cosine}\\
13&d(\tanh(x)) &=& \text{sech}^2(x)~dx &\text{Hyperbolic Tangent}\\
14&d(\coth(x)) &=& -\text{csch}^2(x)~dx &\text{Hyperbolic Cotangent}\\
\end{array}
\]

Answer 13

thank you wio!

Answer 14

The rules we need to use here are 11, 10 and trig rules 1 2

Answer 15

\[
\begin{split}
d(f(x))
&= d(\sqrt 3\sin x+\cos x)&\quad \\
&= d(\sqrt 3\sin x)+d(\cos x)& &\text{ Rule 11} \\
&= \sqrt 3~d(\sin x)+d(\cos x)& &\text{ Rule 10} \\
&= \sqrt 3\cos x~dx+d(\cos x)& &\text{ Trig Rule 1} \\
&= \sqrt 3\cos x~dx-\sin x~dx& &\text{ Trig Rule 2} \\
&= (\sqrt 3\cos x-\sin x)~dx \\
\end{split}
\]This means:\[
\frac{df}{dx} = \sqrt 3\cos x-\sin x
\]

Answer 16

Do you know how to find the roots of this function?