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
2(4-16) - (-30)
= 2(-12) -(-30)
= 24 - (-30) ....Error right here....a negative multiplies a positive = negative (= -24 + 30)
= 24 + 30
= 54
Correct:
2(4-16) - (-30)
= 2(-12) -(-30)
= -24 - (-30)
= -24 + 30
= 6
Use the conjugate of the 1st solution to find the second. which is -3 - i. or a
The ways to illustrate the shapes are given.
<h3>How to depict the shape?</h3>
11. In each new figure, one vertex is introduced, and each vertex is joined to each other vertex.
12. In each new figure, 2 rows and 1 column are introduced, and the shading at the corner moves 90 degrees clockwise.
13. In each new figure, one row and one column are introduced, and exactly half of the rectangles are shaded.
14. In each new figure, except at the bottom vertex, two perpendicular lines are introduced at the remaining vertices.
15. In each new figure, one inverted white triangle is inscribed in the orange triangle.
16. The number of squares increases in the pattern. There are blue circles inscribed in one diagonal square and there are yellow circles inscribed in the remaining squares. Also, the blue circles in the diagonal alternate.
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