find the derivative of $f(x)=sqrt{x^{2}+sqrt{x^{3}+cos(x^{2})}}$.\n$f(x)=\frac{1}{2}x^{2}+sqrt{x^{3}+cos(x^{2…

find the derivative of $f(x)=sqrt{x^{2}+sqrt{x^{3}+cos(x^{2})}}$.\n$f(x)=\frac{1}{2}x^{2}+sqrt{x^{3}+cos(x^{2})}^{-\frac{1}{2}}2x+\frac{1}{2}x^{3}+cos(x^{2})^{-\frac{1}{2}}3x^{2}-sin(x^{2})2x$\nthe function is not differentiable\n$f(x)=\frac{1}{2}x^{2}+sqrt{x^{3}+cos(x^{2})}^{-\frac{1}{2}}2x+\frac{1}{2}3x^{2}-sin(x^{2})2x^{-\frac{1}{2}}$\n$f(x)=\frac{1}{2}x^{2}+sqrt{x^{3}+cos(x^{2})}^{-\frac{1}{2}}2x+\frac{1}{2}x^{3}+cos(x^{2})^{-\frac{1}{2}}$\n$f(x)=\frac{1}{2}x^{2}+(x^{3}+cos(x^{2}))^{\frac{1}{2}}^{-\frac{1}{2}}2x+\frac{1}{2}x^{3}+cos(x^{2})^{-\frac{1}{2}}3x^{2}-sin(x^{2})2x$

find the derivative of $f(x)=sqrt{x^{2}+sqrt{x^{3}+cos(x^{2})}}$.\n$f(x)=\frac{1}{2}x^{2}+sqrt{x^{3}+cos(x^{2})}^{-\frac{1}{2}}2x+\frac{1}{2}x^{3}+cos(x^{2})^{-\frac{1}{2}}3x^{2}-sin(x^{2})2x$\nthe function is not differentiable\n$f(x)=\frac{1}{2}x^{2}+sqrt{x^{3}+cos(x^{2})}^{-\frac{1}{2}}2x+\frac{1}{2}3x^{2}-sin(x^{2})2x^{-\frac{1}{2}}$\n$f(x)=\frac{1}{2}x^{2}+sqrt{x^{3}+cos(x^{2})}^{-\frac{1}{2}}2x+\frac{1}{2}x^{3}+cos(x^{2})^{-\frac{1}{2}}$\n$f(x)=\frac{1}{2}x^{2}+(x^{3}+cos(x^{2}))^{\frac{1}{2}}^{-\frac{1}{2}}2x+\frac{1}{2}x^{3}+cos(x^{2})^{-\frac{1}{2}}3x^{2}-sin(x^{2})2x$

Answer

Explanation:

Step1: Let (u = x^{2}+\sqrt{x^{3}+\cos(x^{2})}), then (f(x)=\sqrt{u}=u^{\frac{1}{2}})

By the chain - rule, (f^{\prime}(x)=\frac{1}{2}u^{-\frac{1}{2}}\cdot u^{\prime})

Step2: Find (u^{\prime})

Let (v = x^{3}+\cos(x^{2})), then (u=x^{2}+\sqrt{v}=x^{2}+v^{\frac{1}{2}}) (u^{\prime}=2x+\frac{1}{2}v^{-\frac{1}{2}}\cdot v^{\prime})

Step3: Find (v^{\prime})

(v = x^{3}+\cos(x^{2})), so (v^{\prime}=3x^{2}-\sin(x^{2})\cdot2x)

Step4: Substitute (v^{\prime}) into (u^{\prime}) and then (u^{\prime}) into (f^{\prime}(x))

(u^{\prime}=2x+\frac{1}{2}(x^{3}+\cos(x^{2}))^{-\frac{1}{2}}(3x^{2}-2x\sin(x^{2}))) (f^{\prime}(x)=\frac{1}{2}[x^{2}+\sqrt{x^{3}+\cos(x^{2})}]^{-\frac{1}{2}}[2x+\frac{1}{2}(x^{3}+\cos(x^{2}))^{-\frac{1}{2}}(3x^{2}-2x\sin(x^{2}))])

Answer:

(f^{\prime}(x)=\frac{1}{2}[x^{2}+\sqrt{x^{3}+\cos(x^{2})}]^{-\frac{1}{2}}[2x+\frac{1}{2}(x^{3}+\cos(x^{2}))^{-\frac{1}{2}}(3x^{2}-2x\sin(x^{2}))]) (First option)