Am I misapplying the chain rule when differentiating $x^{5x+7}$ with respect to $x$?

Question

The problem I am attempting to solve is:

\begin{align} y=x^{5x+7} \\ \text{Find $\frac{dy}{dx}$} \end{align}

Here is my working so far:

$$\begin{align} \text{let }u &= 5x+7 \\ \frac{dy}{dx}&=\frac{dy}{du} \cdot \frac{du}{dx} \\ \frac{dy}{du}&=ux^{u-1}=(5x+7)x^{5x+6} \\ \frac{du}{dx}&=5 \\ \therefore \frac{dy}{dx}&=(25x+35)x^{5x+6} \end{align}$$

(Additionally, I'm having trouble using the align environment in Mathjax. This question wasn't formatted very well. If someone could give me some help in the comments, then I would be very thankful.)

Answer 1

If $u= 5x+7$ and $y = x^{5x+7} = x^u$ then if you attempt to solve $\frac {dy}{dx}=\frac{dy}{du}\frac {du}{dx}$ then you must solve $\frac {dy}{du} = \frac {dx^u}{du}$ but $x$ is dependent on $u$ so you'd have to solve $\frac {dx^u}{du} = \frac {dx^u}{dx}\frac {dx}{du}$ but to solve $\frac {dx}{du}$ we have $x$ is dependant on $u$ and .... and we'd have an infinite loop.

(By the way, if we treat $x$ as constant we'd have $\frac {dx^u}{du}=\ln(x)x^u$; not $\frac {dx^u}{dx} = ux^{u-1}$ which is what we'd have if we treated $u$ as a constant and differentiating in respect to $x$.)

To do the properly we need to express $y =x^{5x+7}$ entirely in terms of $u$

So if $u = 5x +7$ then $x = \frac {u-7}5$ and so $y =x^{5x+7} = x^u = (\frac {u-1}5)^u$ which ..... just doesn't help.

Better we do: let $y = x^{5x + 7} = e^{\ln x(5x+7)}$ the let $u = \ln x(5x+7)$ so $y=e^u$ and we can do

$\frac {dy}{dx} = \frac {dy}{du}\frac {du}{dx}$.

$\frac {dy}{du}=\frac{de^u}{d^u} = e^u= y$ was easy enough. And we can use the product rule to figure $\frac {du}{dx}$.

Let $w= \ln x$ and $v = 5x+7$ then $\frac {du}{dx} =\frac {dv\cdot w}{dx} = \frac{dv}{dx} w + \frac {dw}{dx}v$.

And $\frac {dv}{dx} = \frac {d(5x+7)}{dx}= 5$ and $\frac {dw}{dx}= \frac {d\ln x}{dx} = \frac 1x$ so

$\frac {du}{dx} = (5\ln x + \frac 1x(5x + 7))=(5\ln x + 5 +\frac 7x)$ and do

$\frac {dy}{dx}=\frac {dy}{du}\frac {du}{dx} = \frac {dy}{du}(\frac {dv}{dx}w + \frac {dw}{dx}v) = e^u(5\ln x + 5 +\frac 7x)= e^{5x+7}(5\ln x + 5 + \frac 7x)$.

Answer 2

It is $$\ln(y)=(5x+7)\ln(x)$$ so $$\frac{y'}{y}=5\ln(x)+\frac{5x+7}{x}$$

Answer 3

You forgot to put the x in the base of the exponent in terms of $u$.
$$x^{(5x+7)} = e^{\ln(x)(5x+7)}$$ Let $u=\ln(x)(5x+7)$ $$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} $$ $$\frac{dy}{du} = (e^u)’ = e^u$$ $$ \frac{du}{dx} = (\ln(x)(5x+7))’= \frac{5x+7}{x} + 5ln(x)$$ $$\frac{dy}{dx} = x^{(5x+7)}(5+\frac{7}{x}+ 5\ln(x))$$

Answer 4

In your function $f(x):=x^{5x+7}$ a two-variables function is hidden, namely $$P(u,v):= u^v\ .$$ Therefore we have the total situation $$f(x):=P\bigl(u(x), v(x)\bigr),\qquad u(x):=x,\quad v(x):=5x+7\ .$$ The chain rule then says that $$\eqalign{f'(x)&=P_u\bigl(u(x), v(x)\bigr)\cdot u'(x)+P_v\bigl(u(x), v(x)\bigr)\cdot v'(x)\cr &=v\, u^{v-1}\cdot1+u^v\,\log u\cdot 5\cr &=(5x+7)x^{5x+6}+5x^{5x+7}\log x\ .\cr}$$ (Note that the product rule for $ {d\over dx}\bigl(f(x)\cdot g(x)\bigr)$ also comes from a hidden two-variables function, namely $M(u,v):=u\cdot v$.)

Answer 5

$$y=x^{5x+7} \implies \ln y=(5x+7)\ln x$$

$$\frac{d}{dx}\ln y=\frac{d}{dx}(5x+7)\ln x$$ Using product rule for differentiation, $$\frac1y\frac{dy}{dx}=(5x+7)\frac{d}{dx}\ln x+\ln x\frac{d}{dx}(5x+7)$$ $$\frac1y\frac{dy}{dx}=\frac{5x+7}{x}+\ln x(5)$$ $$\frac{dy}{dx}=\frac yx(5x+5x\ln x+7)$$ $$\color{blue}{\frac{dy}{dx}=x^{5x+6}(5x+5x\ln x+7)}$$