Question

If cos z = u + v and u = sin x, v = cos x find the total derivative of
z with respect to x​

Answer 1

$\textbf{Given:}$

$cosz=u+v\;\text{and}\;u=sinx,\;v=cosx$

$\textbf{To find:}$

$\text{Total derivative of z with respect to x}$

$\textbf{Solution:}$

$\text{Consider,}$

$cosz=u+v$

$cosz=sinx+cosx$

$\text{Differentiate with respect to x}$

$\dfrac{d(cosz)}{dx}=\dfrac{d(sinx)}{dx}+\dfrac{d(cosx)}{dx}$

$sinz\,\dfrac{dz}{dx}=cosx+(-sinx)$

$sinz\,\dfrac{dz}{dx}=cosx-sinx$

$\dfrac{dz}{dx}=\dfrac{cosx-sinx}{sinz}$

$\dfrac{dz}{dx}=\dfrac{cosx-sinx}{\sqrt{1-cos^2z}}$

$\implies\dfrac{dz}{dx}=\dfrac{cosx-sinx}{\sqrt{1-(sinx+cosx)^2}}$

$\textbf{Answer:}$

$\boxed{\bf\dfrac{dz}{dx}=\dfrac{cosx-sinx}{\sqrt{1-(sinx+cosx)^2}}}$

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