Question

Find from first principles the derivative of the sin x²

Answer 1

dy/dx(sinx2) = (cosx^2)(2x)

Answer 2

The derivative of the sin x² using first principles is $2\times x\times cos(x^2)$.

Step-by-step explanation:

Given:

Using first principle, derivative of the $sin x^2$ is,

$\lim_{t \to 0} \frac{sin((x+t)^2-sin(x)^2)}{t}$

= $\lim_{t \to 0} \frac{2sin(\frac{(x+t)^2-x^2}{2})cos(\frac{(x+t)^2+x^2}{2})}{t}$

Multiply and divide by $\frac{2x+t}{2}$, we get

= $\lim_{t \to 0} \frac{2sin(\frac{2xt+t^2}{2})cos(\frac{((x+t)^2+x^2)}{2} )}{t} \times\frac{\frac{2x+t}{2} }{\frac{2x+t}{2} }$

=$\lim_{t \to 0}2\frac{sin(\frac{2xt+t^2}{2})}{\frac{2xt+t^2}{2}} \times\frac{\frac{2x+t}{2} }{1} \times cos\frac{(x+t)^2+x^2}{2}$

=$\lim_{t \to \0} 2\times 1\times \frac{\frac{2x+t}{2} }{1} \times cos\frac{(x+t)^2+x^2}{2}$

=$2\times \frac{2x+0}{2}\times cos\frac{(x+0)^2+x^2}{2}$

=$2\times x\times cos(x^2)$