Question

Suppose $f'(x) = \sqrt{1 + 5x^4}$. $\frac{d}{dx}f(2x) = 2 + 160x^4$ $\frac{d}{dx}f(4x^4) = 16x^3 + 5120x^{19}$

          Suppose $f'(x) = \sqrt{1 + 5x^4}$.
$\frac{d}{dx}f(2x) = 2 + 160x^4$
$\frac{d}{dx}f(4x^4) = 16x^3 + 5120x^{19}$

Suppose f'(x) = √(1 + 5x^4).
(d)/(dx)f(2x) = 2 + 160x^4
(d)/(dx)f(4x^4) = 16x^3 + 5120x^19

Added by Jose Miguel C.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Sri K

03/16/2023

Step 1

3 Given y=4x^2, when x=4, y=4(4)^2=64 Now, when x=4 and Δx=0.3, we can calculate the new y value: y_new = 4(4+0.3)^2 = 4(4.3)^2 = 4(18.49) = 73.96 Therefore, the change in y, Δy = y_new - y = 73.96 - 64 = 9.96 Show more…

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Let y=4x^(2). Find the change in y,Delta y when x=4 and Delta x=0.3 Find the differential dy when x=4 and dx=0.3 Suppose f^(')(x)=sqrt(1+5x^(4)) (d)/(dx)f(2x)= (d)/(dx)f(4x^(4))= Suppose f')=V1+5x4 dx f(2x)=2+160x4 f(4x4) 16x3+5120x19 x