Question

Let f(1) = 3, f'(1) = `-1/3`, g(1) = −4 and g'(1) = `-8/3`. The derivative of `sqrt([f(x)]^2 + [g(x)]^2` w.r.r. x at x = 1 is ______.

Options

• `-29/25`

• `7/3`

• `31/15`

• `29/15`

Answer 1

Let f(1) = 3, f'(1) = `-1/3`, g(1) = −4 and g'(1) = `-8/3`. The derivative of `sqrt([f(x)]^2 + [g(x)]^2` w.r.t. x at x = 1 is `bbunderline(29/15)`.

Explanation:

Let y = `sqrt([f(x)]^2 + [g(x)]^2`

Then `dy/dx = 1/(2 sqrt([f(x)]^2 + [g(x)]^2)) * [2f(x)*f'(x) + 2g(x)*g'(x)]`

∴ `(dy/dx)_("at" x = 1) = 1/(2 sqrt([f(1)]^2 + [g(1)]^2)) * [2f(1)*f'(1) + 2g(1)*g'(1)]`

= `1/(2sqrt(9 + 16)) xx [2(3)(-1/3) + 2(-4)(-8/3)]`

= `1/(2sqrt25) xx [-2 + 64/3]`

= `1/(2 xx 5) xx -6/3 + 64/3`

= `1/10 xx (64 - 6)/3`

= `1/10 xx 58/3`

= `58/30`

= `29/15`