Question

Find the derivatives of sin(5x –8) with respect to first principle

Answer 1

answer of ur question

Answer 2

Derivatives of $\sin (5 x-8) \text { is } \bold{5 \cos (5 x-8)}$

Given:

$\sin (5 x-8)$

To find:

Derivatives of  $\sin (5 x-8)$

Solution:

The derivative of the first principle is a way to find general expression of a “slope” of a curve using algebraic equation. The process is also known as delta method. The derivative is also used to find rate of change for an instance.  

$\begin{array}{l}{f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}}\\ \\ {f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\sin (5(x+h)-8)-(\sin (5 x-8))}{h}}\end{array}$

Using formula,  

$\sin C-\sin D=\frac{2 \cos (c+D)}{2} \cdot \frac{\sin (c-D)}{2}$

Let us take C as $\sin (5(x+h)-8$  and D as $\sin (5 x-8)$  we get  $\frac{2 \cos \left(5 h+\frac{5 x}{2}-8\right)}{2} \cdot \frac{\sin \left(\frac{5 h}{2}\right)}{2}$

$\lim _{h \rightarrow 0} \frac{2 \cos \left(5 h+\frac{5 x}{2}-8\right)}{2} \cdot \frac{\sin \left(\frac{5 h}{2}\right)}{2}=2 \cos (5 x-8) \cdot \frac{5}{2}=5 \cos (5 x-8)$

Therefore, the derivative of $\sin (5 x-8)\ \text{is}\ 5 \cos (5 x-8)$