Question

If $f\left(x\right)=(x^2-1)(3+x^2{)}^{\frac{1}{2}}(8+x^4{)}^{\frac{1}{2}},$ then $f^{\prime }\left(1\right)$ is equal to

• A. $0$
• B. $3$
• C. $12$
• D. $7$

Answer 1

The correct option is C $12$

Let differentiation of $(3+x^2{)}^{\frac{1}{2}}=K$
and differentiation of $(8+x^4{)}^{\frac{1}{2}}=L$
$\Rightarrow f^{\prime }\left(x\right)=(x^2-1)\left(K\right)(8+x^4{)}^{\frac{1}{2}}+(x^2-1\left)\right(3+x^2{)}^{\frac{1}{2}}\left(L\right)+\left(2x\right)(3+x^2{)}^{\frac{1}{2}}(8+x^4{)}^{\frac{1}{2}}$
$\Rightarrow f^{\prime }\left(1\right)=0+0+2\left(4{)}^{\frac{1}{2}}\right(8+1{)}^{\frac{1}{2}}$
$f^{\prime }\left(1\right)=2\cdot 2\cdot 3=12$