I don't understand how to take the derivative of this expression. $$y=2^{\sin (\pi x)}$$
$\endgroup$6 Answers
$\begingroup$The function is $e^{(\ln 2)(\sin(\pi x))}$. Now use the Chain Rule.
Remark: Essentially the same idea will deal with the derivative of $(f(x))^{g(x)}$, where $f(x)$ is a positive function. Use the fact that $$(f(x))^{g(x)}=e^{(\ln(f(x))(g(x))},$$ and use the Chain Rule.
$\endgroup$ $\begingroup$You'll have to apply the chain rule.
$\endgroup$ $\begingroup$Let $f(x)=2^{\sin(\pi x)}$ and let $g(x)=\ln(f(x))$. Then $g(x)=\ln(2)\cdot\sin(\pi x)$ so $$ g^\prime(x)=\ln(2)\pi\cdot\cos(\pi x)\tag{1} $$ But we also have $$ g^\prime(x)=\frac{f^\prime(x)}{f(x)} $$ which gives $$ f^\prime(x)=f(x)\cdot g^\prime(x)\tag{2} $$ Putting (1) and (2) together gives $$ f^\prime(x)=2^{\sin(\pi x)}\cdot \ln(2)\pi\cdot\cos(\pi x) $$
$\endgroup$ $\begingroup$Let the equation be $$a^{u(x)}=e^{u(x)\log(a)}$$ So the derivative is $$\Big(u(x)\log(a)\Big)'e^{u(x)\log(a)}=\log(a)~u'(x)~e^{u(x)\log(a)}=\log(a)~u'(x)~a^{u(x)}$$
$\endgroup$ $\begingroup$The most important thing you have to use here is the Chain Rule which states that: $$(f(g(x)))' = f'(g(x)) \cdot g'(x)$$
$$ y = 2^{\sin (\pi x)} \implies \log_2y = \sin (\pi x)\\ \implies \frac{\ln(y)}{\ln(2)}=\sin(\pi x)\\ \implies \frac{1}{y} \frac{dy}{dx} = \ln(2)\cdot\cos(\pi x)\cdot \pi \\ \implies \frac{dy}{dx} = \pi\ln(2)\cdot 2^{sin (\pi x)}\cdot\cos (\pi x)$$ $\endgroup$ $\begingroup$
The trick is to convert the expresssion to a power of $e$. $$e^{\ln 2}=e^{\ln 2}$$ $$\text{so }2=e^{\ln 2}$$ Then $$2^{\sin (\pi x)}=\bigg(e^{\ln 2}\bigg) ^{\sin(\pi x)}=e^{\ln 2\cdot\sin(\pi x)}$$ Now just apply the chain rule. $$\frac{de^{\ln 2\cdot\sin(\pi x)}}{d \ln 2\cdot\sin(\pi x)}\cdot\frac{d \ln 2\cdot\sin(\pi x)}{d\sin(\pi x)}\cdot\frac{d\sin(\pi x)}{d\pi x}\cdot\frac{d\pi x}{dx}$$
If you prefer, another way of writing this out is with multiple variables. given$\dots$ $$w=\pi x$$ $$v=\sin w$$ $$u=(\ln2)\cdot v$$ the derivative expression becomes $$\frac{de^{u}}{du}\cdot\frac{du}{dv}\cdot\frac{dv}{dw}\cdot\frac{dw}{dx}$$
$\endgroup$
