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Use part 1 of the fundamental theorem of calculus to find the derivative of the function. y = cos x...
4 months ago
Q:
Use part 1 of the fundamental theorem of calculus to find the derivative of the function. y = cos x (3 + v5)8 dv sin x
Accepted Solution
A:
It looks like
[tex]y(x)=\displaystyle\int_{\cos x}^{\sin x}(3+v^5)^8\,\mathrm dv[/tex]
(If the limits are in the wrong order, just multiply the result by -1)
Split the integral at an arbitrary value between [tex]\cos x[/tex] and [tex]\sin x[/tex], and write [tex]y(x)[/tex] as
[tex]y(x)=\displaystyle\left\{\int_{\cos x}^c+\int_c^{\sin x}\right\}(3+v^5)^8\,\mathrm dv[/tex]
[tex]y(x)=\displaystyle\int_c^{\sin x}(3+v^5)^8\,\mathrm dv-\int_c^{\cos x}(3+v^5)^8\,\mathrm dv[/tex]
Then by the FTC,
[tex]\dfrac{\mathrm dy}{\mathrm dx}=(3+\sin^5x)^8\cdot\dfrac{\mathrm d\sin x}{\mathrm dx}-(3+\cos^5x)^8\cdot\dfrac{\mathrm d\cos x}{\mathrm dx}[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\cos x(3+\sin^5x)^8+\sin x(3+\cos^5x)^8[/tex]