Question

Let f and g be differentiable function such that $f^{\prime }\left(x\right)=2g\left(x\right)$ and $g^{\prime }\left(x\right)=-f\left(x\right)$, and let $T\left(x\right)={\left(f\left(x\right)\right)}^2-{\left(g\left(x\right)\right)}^2$. Then $T^{\prime }\left(x\right)$ is equal to

• A. $T\left(x\right)$
• B. $0$
• C. $2f\left(x\right)g\left(x\right)$
• D. $6f\left(x\right)g\left(x\right)$

Answer 1

The correct option is D $6f\left(x\right)g\left(x\right)$

The given equation is:

$T\left(x\right)=f(x{)}^2-g(x{)}^2$

$\Rightarrow T\left(x\right)=\left(f\right(x)-g(x\left)\right)\left(f\right(x)+g(x\left)\right)$

Differentiating once w.r.t to x we get,

$\Rightarrow T^{\prime }\left(x\right)=\left(f^{\prime }\right(x)+g^{\prime }(x\left)\right)\left(f\right(x)-g(x\left)\right)+\left(f^{\prime }\right(x)-g^{\prime }(x\left)\right)\left(f\right(x)+g(x\left)\right)$

$\Rightarrow T^{\prime }\left(x\right)=\left(2g\right(x)-f(x\left)\right)\left(f\right(x)-g(x\left)\right)+\left(2g\right(x)+f(x\left)\right)\left(f\right(x)+g(x\left)\right)$

$\Rightarrow T^{\prime }\left(x\right)=2g\left(x\right)f\left(x\right)-2g(x{)}^2-f(x{)}^2+f\left(x\right)g\left(x\right)+2g\left(x\right)f\left(x\right)+2g(x{)}^2+f(x{)}^2+f\left(x\right)g\left(x\right)$

$\Rightarrow T^{\prime }\left(x\right)=6f\left(x\right)g\left(x\right)$ .....Answer