Question

Find $\frac{dx}{dx}$, if $x^5{y}^7=(x+y{)}^1$.

Answer 1

We have,

$x^5{y}^7=x+y$

On differentiating w,r,t $x$, we get

$\Rightarrow \frac{d}{dx}\left(x^5{y}^7\right)=\frac{d}{dx}(x+y)$

$\Rightarrow x^5\times 7y^6\frac{dy}{dx}+y^7\times 5x^4=1+\frac{dy}{dx}$

$\Rightarrow (7x^5{y}^6-1)\frac{dy}{dx}=1-5x^4{y}^7$

$\Rightarrow \frac{dy}{dx}=\frac{(1-5x^4{y}^7)}{(7x^5{y}^6-1)}$

Hence, this is the answer.