Question

how to find the derivative of log (2x+5) by method of first principle​

Answer 1

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Answer 2

Concept

The term "derivative" refers to the use of mathematics to find a general expression for a curve's slope. The delta approach is another name for it. The derivative is a measure of the rate of change at the present moment, which is equal to

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

Given

The given expression is log (2x+5).

Find:

We have to find the derivative by the method of the first principle​.

Solution:

Apply the first principle​ method

$\begin{aligned}&f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\\&=\lim _{h \rightarrow 0} \frac{\log (2 x+2 h+5)-\log (2 x+5)}{h}\\&=\lim _{h \rightarrow 0} \frac{1}{h} \log \left(\frac{2 x+2 h+5}{2 x+5}\right)\\&=\lim _{h \rightarrow 0} \frac{1}{h} \log \left(\frac{2 x+5+2 h}{2 x+5}\right)\\&=\lim _{h \rightarrow 0} \frac{1}{h} \log \left(1+\frac{2 h}{2 x+5}\right)\\\end{aligned}$

Simplify further

$\begin{aligned}&=\lim _{h \rightarrow 0} \log \left(1+\frac{2 h}{2 x+5}\right)^{\frac{1}{h}}\\&=\lim _{h \rightarrow 0} \log \left[\left(1+\frac{2 h}{2 x+5}\right)^{\frac{2 x+5}{2 h}}\right]^{\frac{2}{2 x+5}}\\&=\log \mathrm{e}^{\frac{2}{2 x+5}} \quad \cdots\left[\because \lim _{x \rightarrow 0}(1+p x)^{\frac{1}{\beta r}}=\mathrm{e}\right]\\&=\frac{2}{2 x+5} \log \mathrm{e}\\&=\frac{2}{2 x+5} \quad \cdots[\because \log e=1]\end{aligned}$

Hence the derivative is  $\frac{2}{2 x+5}$ .