Question

The derivative of $(x^3+\ln 6)\left(\sqrt{x}\right)$ will be

• A. $\frac{3}{2}{x}^{3/2}$
• B. $3x^2\sqrt{x}$
• C. $3x^2\sqrt{x}+\frac{x^3+\ln 6}{2\sqrt{x}}$
• D. None of the above

Answer 1

The correct option is C $3x^2\sqrt{x}+\frac{x^3+\ln 6}{2\sqrt{x}}$

If we closely watch the given function, it is the product of two different functions i.e., $(x^3+\ln 6)$ and $\sqrt{x}$. So don't you think that we can solve this by applying product rule for derivatives? The rule says that,
$\frac{d}{d\left(x\right)}\left[f\right(x)\times g(x\left)\right]=\frac{d}{d\left(x\right)}\left(f\right(x\left)\right)\times g\left(x\right))+\frac{d}{dx}\left(g\right(x\left)\right)\times f(x\left)\right)So,\frac{d}{d\left(x\right)}\left[\right(x^3+\ln 6\left)\right(\sqrt{x}\left)\right]=\frac{d}{d\left(x\right)}(x^3+\ln 6)\times \left(\sqrt{x}\right)+\frac{d}{d\left(x\right)}\left(\sqrt{x}\right)\times (x^3+ln6)=(3x^2+0)\left(\sqrt{x}\right)+(x^3+\ln 6)\left(\frac{1}{2}{x}^{-1/2}\right)=3x^2\sqrt{x}+\frac{x^3+\ln 6}{2\sqrt{x}}$
Which is the option C.
By the way this question can also be solved using addition rule for differentiation if you write the above function as,
$x^2\sqrt{x}+\ln 6\sqrt{x}$
i.e. $x^{5/2}+\ln 6\sqrt{x}$