If $k(x)=g(x)\cdot f'(7x)$, what is the slope of the tangent line at $x=1$?

Key concept: This is a product of two functions, and one factor contains a chain-rule composition. The tangent slope is $k'(1)$.

Step by step

Step 1: Differentiate $k(x)$

Given

$k(x)=g(x)\cdot f'(7x)$

Use the product rule:

$k'(x)=g'(x)\,f'(7x)+g(x)\cdot \frac{d}{dx}[f'(7x)]$

Now apply the chain rule to $f'(7x)$:

$\frac{d}{dx}[f'(7x)]=7f''(7x)$

So,

$k'(x)=g'(x)f'(7x)+7g(x)f''(7x)$

Step 2: Evaluate at $x=1$

$k'(1)=g'(1)f'(7)+7g(1)f''(7)$

Final answer

The slope of the tangent line at $x=1$ is

$\boxed{g'(1)f'(7)+7g(1)f''(7)}$

If numerical values for $g(1)$, $g'(1)$, $f'(7)$, and $f''(7)$ are given elsewhere, substitute them into this formula.

Pitfall alert

A common mistake is to differentiate $f'(7x)$ as just $f''(7x)$. The inner function $7x$ contributes an extra factor of $7$.

Try different conditions

If the problem had $k(x)=g(x)f(7x)$ instead, the derivative would be $k'(x)=g'(x)f(7x)+7g(x)f'(7x)$. The structure is similar, but the derivative of the second factor changes.

Further reading

product rule, chain rule, tangent line