Where did it come from? (derivative of exponential)

Question

We all know this rule:

$\text{If } y = a^{f(x)} \text{ then } y' = a^{f(x)} \: f'(x)\ln a$

In my book there is the example:

Find $\frac{d}{dx}\left((x^{2} + 1)^{\sin x}\right)$

According to the rule, my answer is:

$(x^{2} + 1)^{\sin x} \cdot \cos x \cdot \ln(x^{2} + 1)$

But the answer of the book was:

$$(x^{2} + 1)^{\sin x}\left(\frac{2x \sin x}{x^{2} + 1} + \cos x \cdot \ln(x^{2} + 1)\right)$$

So, where did $\frac{2x \sin x}{x^{2} + 1}$ come from?

Answer 1

I will assume that this is not homework and go ahead to give a full solution.

As others have said, your rule only works for constant $a$. Following the hint in danielson's answer and setting $y=f(x)$, you have $$\log y=\sin x\log\left(1+x^2\right)$$ and differentiating both sides with respect to $x$ and using the chain and product rules gives $$\frac1y\frac{dy}{dx}=\cos x\log\left(1+x^2\right)+\frac{2x\sin x}{1+x^2}\quad\text{so that}\quad \frac{dy}{dx}=y\left(\cos x\log\left(1+x^2\right)+\frac{2x\sin x}{1+x^2}\right).$$Substituting back for $y$ gives the result.

Tell me if you don't understand any of the applications of the chain rule.

Answer 2

The rule that you quote is only valid if a is a constant.

Answer 3

The rule that you have given is for $\textit{constant}$ $a$, while the base of your function is a $\textit{function}$. So the rule does not apply here.

To fix this, let $f(x) = (x^2 + 1)^{\sin(x)}$ and consider $\ln(f(x))$. Differentiate the new function, while also utilizing a nice property of exponents inside logarithmic functions, and try to deduce the derivative of the original function from this.

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Answer 4

Start with $$f(x)=e^{\ln(x^2+1)\sin(x)}$$

The chain rule gives

$$f'(x)=e^{\ln(x^2+1)\sin(x)}\times (\ln(x^2+1)\sin(x))'=(x^2+1)^{\sin(x)}(\ln(x^2+1)\sin(x))'$$

Take it from here.

Answer 5

The rule is $$\frac d{dx}u(x)^{v(x)}=v(x)u(x)^{v(x)-1}\frac{du}{dx}+u(x)^{v(x)}(\ln u(x))\frac{dv}{dx}.$$ You can derive this by the method of logarithmic differentiation. It is a special case of the two-variable chain rule: $$\frac{dw}{dx}=\frac{\partial w}{\partial u}\frac{du}{dx}+\frac{\partial w}{\partial v}\frac{dv}{dx}.$$ Notice how it simplifies to familiar rules if either $u(x)$ or $v(u)$ is a constant. This makes the general formula easy to remember. If $u$ is variable and $v$ is constant then, $$(u^v) '=vu^{v-1}u';$$ if $v$ is variable and $u$ is constant, then $$(u^v)'=u^v(\ln u)v';$$ if $u$ and $v$ are both variables, then $$(u^v)'=vu^{v-1}u'+u^v(\ln u)v'.$$