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{:[h(x)=arcsin(4x)],[h^(')(x)=?]:}
Choose 1 answer:
(A) 
(1)/(sqrt(1-16x^(2)))
(B) 
(4)/(sqrt(1-16x^(2)))
(C) 
(1)/(sqrt(1+16x^(2)))
(D) 
(4)/(sqrt(1+16x^(2)))

h(x)=arcsin(4x)h(x)=? \begin{array}{l} h(x)=\arcsin (4 x) \\ h^{\prime}(x)=? \end{array} \newlineChoose 11 answer:\newline(A) 1116x2 \frac{1}{\sqrt{1-16 x^{2}}} \newline(B) 4116x2 \frac{4}{\sqrt{1-16 x^{2}}} \newline(C) 11+16x2 \frac{1}{\sqrt{1+16 x^{2}}} \newline(D) 41+16x2 \frac{4}{\sqrt{1+16 x^{2}}}

Full solution

Q. h(x)=arcsin(4x)h(x)=? \begin{array}{l} h(x)=\arcsin (4 x) \\ h^{\prime}(x)=? \end{array} \newlineChoose 11 answer:\newline(A) 1116x2 \frac{1}{\sqrt{1-16 x^{2}}} \newline(B) 4116x2 \frac{4}{\sqrt{1-16 x^{2}}} \newline(C) 11+16x2 \frac{1}{\sqrt{1+16 x^{2}}} \newline(D) 41+16x2 \frac{4}{\sqrt{1+16 x^{2}}}
  1. Apply Chain Rule: To find the derivative of h(x)=arcsin(4x)h(x) = \arcsin(4x), we use the chain rule.
  2. Derivative of arcsin(u): The derivative of arcsin(u)\arcsin(u) with respect to uu is 11u2\frac{1}{\sqrt{1-u^2}}. Here, u=4xu = 4x.
  3. Calculate dudx\frac{du}{dx}: Using the chain rule, h(x)=ddx(arcsin(4x))=ddx(arcsin(u))dudxh'(x) = \frac{d}{dx}(\arcsin(4x)) = \frac{d}{dx}(\arcsin(u)) \cdot \frac{du}{dx}.
  4. Plug in Derivatives: We calculate dudx\frac{du}{dx} where u=4xu = 4x. So, dudx=4\frac{du}{dx} = 4.
  5. Simplify Expression: Now, plug in the derivative of uu and the derivative of arcsin(u)\arcsin(u) into the chain rule formula.\newlineh(x)=11(4x)2×4h'(x) = \frac{1}{\sqrt{1-(4x)^2}} \times 4.
  6. Simplify Expression: Now, plug in the derivative of uu and the derivative of arcsin(u)\arcsin(u) into the chain rule formula.\newlineh(x)=11(4x)2×4h'(x) = \frac{1}{\sqrt{1-(4x)^2}} \times 4. Simplify the expression to get the final derivative.\newlineh(x)=4116x2h'(x) = \frac{4}{\sqrt{1-16x^2}}.

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