Solveeit Logo

Question

Question: How do you find the derivative of \(\ln (4x)\)?...

How do you find the derivative of ln(4x)\ln (4x)?

Explanation

Solution

Start differentiating the expression in the question by using the formula ddx(ln(x))=1x\dfrac{d}{{dx}}(\ln (x)) = \dfrac{1}{x}. As the term involved is 4x, differentiate that with respect to x too and simplify this expression.

Complete Step by Step Solution:
We have to find the derivative of ln(4x)\ln (4x).
Let y=ln(4x)y = \ln (4x) .
Differentiating both sides of the equation with respect to x
dydx=14xddx(4x)\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{4x}}\dfrac{d}{{dx}}(4x)
((\because the derivative of ln(x)=1x\ln (x) = \dfrac{1}{x}.
However, an important thing to be taken into consideration is - the term involved is 4x. So we must also differentiate 4x with respect to x.)
dydx=14x×4\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{4x}} \times 4
Simplifying
dydx=1x\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{x}

dydx=1x\therefore \dfrac{{dy}}{{dx}} = \dfrac{1}{x} is the derivative of ln(4x)\ln (4x).

Note:
Alternate method:
We have to find the derivative of ln(4x)\ln (4x).
Let y=ln(4x)y = \ln (4x).
Let 4x=t4x = t
Differentiating both sides of the equation with respect to x,
4ddx(x)=dtdx\Rightarrow 4\dfrac{d}{{dx}}(x) = \dfrac{{dt}}{{dx}}
dtdx=4\Rightarrow \dfrac{{dt}}{{dx}} = 4
Let dtdx=4\dfrac{{dt}}{{dx}} = 4 be labelled as equation 1.
Substituting the value of 4x in the expression, y=ln(4x)y = \ln (4x) we get y=ln(t)y = \ln (t)
Differentiating both sides of the above equation with respect to x,
dydx=1tdtdx\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{t}\dfrac{{dt}}{{dx}}
((\because the derivative of ln(t)=1t\ln (t) = \dfrac{1}{t}.
As the equation involving the variable t is differentiated with respect to x, we need to differentiate t with respect to x too. Thus we obtain the derivative.)
Substituting the value of dtdx\dfrac{{dt}}{{dx}} obtained in equation 1 in the above expression, we get dydx=1t(4)\dfrac{{dy}}{{dx}} = \dfrac{1}{t}(4).
Now we replace the value of t with 4x and change the variable terms back to x.
dydx=44x\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{4}{{4x}}
dydx=1x\therefore \dfrac{{dy}}{{dx}} = \dfrac{1}{x}