Differentiate y= exp(cos^2(x)+sin^2(x)) by using the chain rule.
First of all instead ,we'll define the chain rule , thus y can be rewritten as y = f (g(x)) , where f(x) = exp (x) and g(x) = cos^2(x) + sin^2(x). Therefore let y = f(u) , dy/dx = dy/du * du/dx , which then gives us dy/dx = exp(cos^2(x)+sin^2(x))du/dx. To find du/dx , we'll use the product rule on both cos^2(x) and sin^2(x) , where g(x)=z(x)h(x) therefore dg/dx = dz/dxh+z*dh/dx. The value of du/dx = 0 , therefore dy/dx =0 . We can check the result if we were to use trigonometric identities , we would find that cos^2(x)+sin^2(x) = 1 , therefore y = exp(1) and dy/dx = 0 .
AJ
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