Use the product rule to differentiate the first function:
![g(x)=f(x)\sin x\implies g'(x)=f'(x)\sin x+f(x)\cos x](https://tex.z-dn.net/?f=g%28x%29%3Df%28x%29%5Csin%20x%5Cimplies%20g%27%28x%29%3Df%27%28x%29%5Csin%20x%2Bf%28x%29%5Ccos%20x)
which means
![g'\left(\dfrac\pi3\right)=f'\left(\dfrac\pi3\right)\sin\dfrac\pi3+f\left(\dfrac\pi3\right)\cos\dfrac\pi3=2-\sqrt3](https://tex.z-dn.net/?f=g%27%5Cleft%28%5Cdfrac%5Cpi3%5Cright%29%3Df%27%5Cleft%28%5Cdfrac%5Cpi3%5Cright%29%5Csin%5Cdfrac%5Cpi3%2Bf%5Cleft%28%5Cdfrac%5Cpi3%5Cright%29%5Ccos%5Cdfrac%5Cpi3%3D2-%5Csqrt3)
Now use quotient rule for the second one.
![h(x)=\dfrac{\cos x}{f(x)}\implies h'(x)=\dfrac{-f(x)\sin x-f'(x)\cos x}{f(x)^2}](https://tex.z-dn.net/?f=h%28x%29%3D%5Cdfrac%7B%5Ccos%20x%7D%7Bf%28x%29%7D%5Cimplies%20h%27%28x%29%3D%5Cdfrac%7B-f%28x%29%5Csin%20x-f%27%28x%29%5Ccos%20x%7D%7Bf%28x%29%5E2%7D)
so you have
Answer:
x=40°
Step-by-step explanation:
Firstly, lets look at some things that we know based on this image:
We have a equilateral triangle(The triangle on the left has 3 tick marked on the sides, so they are equal. It also has 3 of the same angle, so it must be equilateral) and a isosceles triangle (There are two tick marks showing that two of the sides are equal length), the measure of each of the equilateral triangle's angles must be 60° each, the measure of these two triangles together must be 360°, and angle x and the unmarked angle must be the same size as this triangle is isosceles.
To solve this, we can set up an equation to solve for x. To do this, we can add up all of the known angles and set it equal to 360.
![(100+60)+60+(60+x)+x=360\\\\160+60+60+2x=360\\\\280+2x=360\\\\2x=80\\\\x=40](https://tex.z-dn.net/?f=%28100%2B60%29%2B60%2B%2860%2Bx%29%2Bx%3D360%5C%5C%5C%5C160%2B60%2B60%2B2x%3D360%5C%5C%5C%5C280%2B2x%3D360%5C%5C%5C%5C2x%3D80%5C%5C%5C%5Cx%3D40)
Answer:
Third option <3, 9>
Step-by-step explanation:
u and v are two vectors and we know the Cartesian components of these vectors.
We must find the sum of u + v.
If we have the Cartesian components of both vectors then the sum of both is done by adding the components of u with the components of v.
![u = \\v =](https://tex.z-dn.net/?f=u%20%3D%20%3C1%2C%203%3E%5C%5Cv%20%3D%20%3C2%2C%206%3E)
![u + v = (1 + 2) i + (3 + 6) j](https://tex.z-dn.net/?f=u%20%2B%20v%20%3D%20%281%20%2B%202%29%20i%20%2B%20%283%20%2B%206%29%20j)
Where i is the vertical component and j is the horizontal component
Then u + v is
![u + v = (3) i + (9) j](https://tex.z-dn.net/?f=u%20%2B%20v%20%3D%20%283%29%20i%20%2B%20%289%29%20j)
![u+v =](https://tex.z-dn.net/?f=u%2Bv%20%3D%20%3C3%2C%209%3E)
Answer:
The question is not complete.
Step-by-step explanation: