To determine

To find:

The derivative of the function gx=∫tan⁡xx212+t4 dt

Answer

g'x=-sec2⁡x2+tan4⁡x+2x2+x8

Explanation

1) Concept:

i) Properties of definite integral:

∫ba fxdx=-∫ab fxdx

2) Fundamental theorem of Calculus, Part 1:

If f is continuous on a, b, then the function g defined by

gx=∫0x ft dt   a≤x≤b

gx is continuous on a,b, and g'x=f(x)

3) Given:

gx=∫tan⁡xx212+t4 dt

4) Calculation:

Rewrite the given integral as (by using hint given)

∫tan⁡xx212+t4 dt=∫tan⁡x012+t4 dt+∫0x212+t4 dt

Now by using concept i) (Property of definite integral),

=-∫0tan⁡x12+t4 dt+∫0x212+t4 dt

Differentiating with respect to x

g'x=ddx-∫0tan⁡x12+t4 dt+ddx∫0x212+t4 dt

By using Fundamental theorem of Calculus, Part 1,

g'x=-12+t4ddtt- 12+t4ddt0+12+t4ddtt- 12+t4ddt0

Substituting the values of limit (t=tan⁡xand 0 in first term t=x2 and 0 in second term),

g'x=-12+tan⁡x4ddx(tan⁡x)- 12+tan⁡x4ddx0+12+x24ddx(x2)- 12+x24ddx0

=-12+tan4⁡xddxtan⁡x+0+12+x8ddxx2- 0

=-12+(tan⁡x)4sec2⁡x+12+x82x

=-sec2⁡x2+tan4⁡x+2x2+x8

Therefore,

g'x=-sec2⁡x2+tan4⁡x+2x2+x8

Conclusion:

Therefore, the derivative the given integral is

-sec2⁡x2+tan4⁡x+2x2+x8