Question

Suppose that $f$ is differentiable function with the property $f(x+y)=f\left(x\right)+f\left(y\right)+x^2{y}^2$ and $\underset{x\to 0}{lim}\frac{f\left(x\right)}{x}=10$ then $f^{\prime }\left(0\right)$ is equal to

Answer 1

The derivative of a function using the first principle is when L.H.D is equal to the R.H.D.

R.H.D. $f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

L.H.D. $f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x-h)-f\left(x\right)}{-h}$

Evaluating the above we get,

R.H.D.

$\Rightarrow f^{\prime }\left(0\right)={lim}_{h\to 0}\frac{f(0+h)-f\left(0\right)}{h}$

$\Rightarrow f^{\prime }\left(0\right)={lim}_{h\to 0}\frac{f\left(0\right)+f\left(h\right)+0.h^2-f\left(0\right)}{h}$

$\Rightarrow f^{\prime }\left(0\right)={lim}_{h\to 0}\frac{f\left(h\right)}{h}$

$\Rightarrow f^{\prime }\left(0\right)=10$

L.H.D.

$\Rightarrow f^{\prime }\left(0\right)={lim}_{h\to 0}\frac{f(0-h)-f\left(x\right)}{-h}$

$\Rightarrow f^{\prime }\left(0\right)={lim}_{h\to 0}\frac{f\left(0\right)+f(-h)+0.h^2-f\left(0\right)}{-h}$

$\Rightarrow f^{\prime }\left(0\right)={lim}_{h\to 0}\frac{f(-h)}{-h}$

$\Rightarrow f^{\prime }\left(0\right)=10$

Thus L.H.D. = R.H.D. = 10 .....Answer