Answer to Question #91411 in Calculus for Fariha

Question #91411
Find the rate of change of h(x)=2 cos⁡〖(3x+tan⁡x)〗 with respect to x.
1
Expert's answer
2019-07-04T10:28:22-0400

The rate of change of y with respect to x, if one has the original function, can be found by taking the derivative of that function. This will measure the rate of change at a specific point.


"h(x)=2 \\cos\u2061(3x+\\tan\u2061(x))"

The derivative of f with respect to x,"h(x)'" is given by



"h'(x)=(2 \\cos\u2061(3x+\\tan\u2061(x)))^{\\prime}= \\\\\n=-2\\sin(3x+\\tan\u2061(x))(3x+\\tan\u2061(x))^{\\prime}=\\\\\n=-2\\sin(3x+\\tan\u2061(x))(3+\\frac{1}{\\cos^2(x)})"

So,

"h'(x)=-2\\sin(3x+\\tan\u2061(x))\\frac{3\\cos^2(x)+1}{\\cos^2(x)}=-\\frac{2\\sin(3x+\\tan\u2061(x))(3\\cos^2(x)+1)}{\\cos^2(x)}"

We found the rate of change of "h(x)=2 \\cos\u2061(3x+\\tan\u2061(x))" with respect to x



"h'(x)=-\\frac{2\\sin(3x+\\tan\u2061(x))(3\\cos^2(x)+1)}{\\cos^2(x)}"





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