Derivative of $x^{\sin x}$

Question

Why $ \sin x.x^{\sin x-1}\cos x$ wrong? Why I cannot treat $\sin x$ just like usual power $x^a = ax^{a-1}$

Because there is a big difference between a constant and the sine function? — Mark, Oct 13 '19 at 16:54
Note that $x^{\sin(x)}=e^{\sin(x)\log(x)}$. Now, use that chain rule and the product rule. — Mark Viola, Oct 13 '19 at 16:55
Let $y=x^{\sin x}$. Take logs, differentiate implicitly and rearrange ... — Donald Splutterwit, Oct 13 '19 at 16:56

score 5 · Accepted Answer · answered Oct 13 '19 at 16:55

5

The proof of $(x^a)^\prime=ax^{a-1}$ assumes $a$ is constant. More generally$$(x^a)^\prime=x^a(a\ln x)^\prime=x^{a-1}(a+xa^\prime\ln x).$$In your case $a=\sin x$, giving the derivative $x^{a-1}(\sin x+x\cos x\ln x)$.

answered Oct 13 '19 at 16:55

J.G.

115,835

2

+1 for answering the actual question of "why is the power rule wrong here?" – Oct 13 '19 at 17:06
$(x^a)^\prime=x^a(a\ln x)^\prime$ what is it? – Dini Oct 13 '19 at 17:06
@Lifeforbetter My calculation answered that. – J.G. Oct 13 '19 at 17:07
How $(x^a)^\prime $become $ x^a(a\ln x)^\prime$? – Dini Oct 13 '19 at 17:12
1

@Lifeforbetter Let $$y = x^a \implies \ln y = a\ln x \implies \frac{dy}{y} = \frac{y'}{y} = (a\ln x)' \implies y' = y(a\ln x)' = x^a(a\ln x)'$$ – 19aksh Oct 13 '19 at 17:22

score 3 · Answer 2 · answered Oct 13 '19 at 16:57

Write $$x^{\sin x}=e^{\ln x^{\sin x}}=e^{\sin x \ln x}$$

Taking the derivative we get $$ e^{\sin x \ln x}(\sin x \ln x)'=e^{\sin x \ln x}\left(\cos x \ln x + \sin x \cdot \frac{1}{x}\right)=x^{\sin x}\left(\cos x \ln x + \sin x \cdot \frac{1}{x}\right)=x^{\sin x -1}(x \cos x \ln x + \sin x) $$

score 2 · Answer 3 · answered Oct 13 '19 at 16:56

2

Writing $$\ln(y)=\sin(x)\ln(x)$$ then we get by the chain rule $$\frac{y'}{y}=\cos(x)\ln(x)+\sin(x)\times \frac{1}{x}$$

answered Oct 13 '19 at 16:56

Dr. Sonnhard Graubner

95,283

score 1 · Answer 4 · answered Oct 13 '19 at 19:11

1

If you treat $\sin(x)$ as a constant, you get $$\sin(x)x^{\sin(x)-1}$$ If you treat the base as a constant, you get $$x^{\sin(x)}\ln(x)\cos(x)$$ If you add these together, you get $$\sin(x)x^{\sin(x)-1}+x^{\sin(x)}\ln(x)\cos(x)$$ which is the actual derivative of $x^{\sin(x)}$. This is a legitimate application of the multivariate chain rule.

answered Oct 13 '19 at 19:11

2'5 9'2

54,717

I also solved this with the same method! but I have some question about that: why this method work? and why it should be "+" between these two derivatives (why add them together why not multiply)? from where this comes? – User Oct 13 '19 at 19:16
2

Consider $g(u,v)=u\sin(v)$ and $D(x)=(x,x)$. Then $f(x)=(g\circ D)(x)$. And the multivariate chain rule says $\frac{df}{dx}(x)=\left.\frac{\partial g}{\partial u}\right\vert_{(u,v)=D(x)}\cdot\frac{\partial D_1}{\partial x}(x)+\left.\frac{\partial g}{\partial v}\right\vert_{(u,v)=D(x)}\cdot\frac{\partial D_2}{\partial x}(x)=\left.\frac{\partial g}{\partial u}+\frac{\partial g}{\partial v}\right\vert_{(u,v)=D(x)}$. It allows you to treat each instance of $x$ as the sole instance of $x$ with everything else constant. But all these contributions towards a rate of change have to be added together. – 2'5 9'2 Oct 13 '19 at 19:30

User · Answer 5 · 2019-10-13T19:12:00.230

0

first consider $x$ as constant then consider $sinx$ as constant:

$[g(x)^{f(x)}]\prime$=[$({x^{f(x)}})^\prime$] + [$({g(x)}^a)^\prime$]= [$f(x)f(x)^\prime{x^{f(x)-1}}$] +[${g(x)}^a$ $ln(g(x))$ $(g(x))^\prime$]

edited Oct 13 '19 at 19:12

answered Oct 13 '19 at 19:06

User

147

Derivative of $x^{\sin x}$

5 Answers5