Question Number 131775 by Raxreedoroid last updated on 08/Feb/21
![Find the derivative of g(x)=∫_(tan x) ^( x^2 ) (1/( (√(2+t^4 )))) dt](https://www.tinkutara.com/question/Q131775.png)
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{derivative}\:\mathrm{of} \\ $$$${g}\left({x}\right)=\int_{\mathrm{tan}\:{x}} ^{\:{x}^{\mathrm{2}} } \frac{\mathrm{1}}{\:\sqrt{\mathrm{2}+{t}^{\mathrm{4}} }}\:{dt} \\ $$
Answered by bemath last updated on 08/Feb/21
![g′(x) = ((d(x^2 ))/((2+(x^2 )^4 )^(1/2) ))−((d(tan x))/( (√(2+tan^4 x)))) g′(x)= ((2x)/( (√(2+x^8 ))))−((sec^2 x)/( (√(2+tan^4 x)))) Formula f(x) = ∫_(u(x)) ^( v(x)) f(t) dt f ′(x) = f(v(x)).v′(x)−f(u(x)).u′(x)](https://www.tinkutara.com/question/Q131783.png)
$$\:\mathrm{g}'\left(\mathrm{x}\right)\:=\:\frac{\mathrm{d}\left(\mathrm{x}^{\mathrm{2}} \right)}{\left(\mathrm{2}+\left(\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{4}} \right)^{\mathrm{1}/\mathrm{2}} }−\frac{\mathrm{d}\left(\mathrm{tan}\:\mathrm{x}\right)}{\:\sqrt{\mathrm{2}+\mathrm{tan}\:^{\mathrm{4}} \mathrm{x}}} \\ $$$$\:\mathrm{g}'\left(\mathrm{x}\right)=\:\frac{\mathrm{2x}}{\:\sqrt{\mathrm{2}+\mathrm{x}^{\mathrm{8}} }}−\frac{\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}}{\:\sqrt{\mathrm{2}+\mathrm{tan}\:^{\mathrm{4}} \mathrm{x}}}\: \\ $$$$\:\mathrm{Formula}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\int_{\mathrm{u}\left(\mathrm{x}\right)} ^{\:\mathrm{v}\left(\mathrm{x}\right)} \mathrm{f}\left(\mathrm{t}\right)\:\mathrm{dt}\: \\ $$$$\:\mathrm{f}\:'\left(\mathrm{x}\right)\:=\:\mathrm{f}\left(\mathrm{v}\left(\mathrm{x}\right)\right).\mathrm{v}'\left(\mathrm{x}\right)−\mathrm{f}\left(\mathrm{u}\left(\mathrm{x}\right)\right).\mathrm{u}'\left(\mathrm{x}\right) \\ $$
Commented by Raxreedoroid last updated on 08/Feb/21
![Thanks sir](https://www.tinkutara.com/question/Q131790.png)
$${Thanks}\:{sir} \\ $$
Answered by mathmax by abdo last updated on 08/Feb/21
![g^′ (x)=((2x)/( (√(2+x^8 ))))−((1+tan^2 x)/( (√(2+tan^4 x))))](https://www.tinkutara.com/question/Q131776.png)
$$\mathrm{g}^{'} \left(\mathrm{x}\right)=\frac{\mathrm{2x}}{\:\sqrt{\mathrm{2}+\mathrm{x}^{\mathrm{8}} }}−\frac{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}}{\:\sqrt{\mathrm{2}+\mathrm{tan}^{\mathrm{4}} \mathrm{x}}} \\ $$
Commented by Raxreedoroid last updated on 08/Feb/21
![Can you show the steps sir](https://www.tinkutara.com/question/Q131779.png)
$${Can}\:{you}\:{show}\:{the}\:{steps}\:{sir} \\ $$