As the title suggests, I need to differentiate $x$ to the power of $y$ with respect to $x$. Not sure how to start. Do I need to take natural log on both sides?
That is: $\dfrac{d}{dx}x^y=?$
Asked
Active
Viewed 5,285 times
0
-
1$\frac{dx^y}{dx} = yx^{y-1}$. Did you mean to say, differentiate with respect to $y$? – Andrew Poelstra Jul 25 '13 at 04:59
2 Answers
2
We need $$\frac{d(x^y)}{dx}$$
One of the ways could be:
Let $f(x)=x^y\implies \ln f(x)=y \cdot \ln x$
Using Chain Rule for the Left hand side, $$\frac{\ln f(x)}{dx}=\frac{\ln f(x)}{d f(x)}\frac{d f(x)}{dx}=\frac{f'(x)}{f(x)}$$
and Product Rule of Differentiation for the Right, $$\frac{d(y \cdot \ln x)}{dx}=y\cdot \frac{d(\ln x)}{dx}+\ln x\cdot \frac{dy}{dx}=\ln x\cdot \frac{dy}{dx}+\frac 1x\cdot y$$
$$\implies \frac{f'(x)}{f(x)}=\ln x\cdot \frac{dy}{dx}+\frac yx$$
$$\implies f'(x)=f(x)\left(\ln x\cdot \frac{dy}{dx}+\frac yx\right)=x^y\left(\ln x\cdot \frac{dy}{dx}+\frac yx\right)$$
lab bhattacharjee
- 274,582
-
Note that this answer assumes that $y$ is a function of $x$. If $y$ is constant, then this reduces to $\frac{dx^y}{dx}=yx^{y-1}$. – Mario Carneiro Jul 25 '13 at 05:25
-
1
As $x^y=(e^{\ln x})^y=e^{y\ln x}$
$$\frac{d(x^y)}{dx}=\frac{d(e^{y\ln x})}{dx}$$
$$=\frac{d(e^{y\ln x})}{d(y\ln x)}\cdot \frac{ d(y\ln x)}{dx}$$ (applying Chain Rule)
$$=e^{y\ln x}\left(y\cdot \frac{d(\ln x)}{dx}+\ln x\cdot\frac{dy}{dx}\right)$$
$$=x^y\left(y\cdot\frac1x+\ln x\cdot\frac{dy}{dx}\right)$$
lab bhattacharjee
- 274,582